Skip to content

Using the Quadratic Formula,
Finding Quadratic Equation Solutions


Along with factoring quadratics, another way to obtain quadratic equation solutions is to use the quadratic formula.


When we have a standard quadratic equation of the form,    ax^{\tt{2}} + bx + c = 0.

We can solve this equation with the following “quadratic formula”.

\color{blue}{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}b \space \pm \space \sqrt{b^{2} \space – \space 4ac}}{2a}}}


The presence of a combined plus and minus sign ‘ \pm‘,  indicates that there are two possible solutions.
So we can break the quadratic formula down a little bit further.

{\footnotesize{x_1 = \space}} {\boldsymbol{\frac{{\text{-}}b \space + \space \sqrt{b^{2} \space – \space 4ac}}{2a}}} \space \space , \space \space {\footnotesize{x_2 = \space}} {\boldsymbol{\frac{{\text{-}}b \space – \space \sqrt{b^{2} \space – \space 4ac}}{2a}}}


Though sometimes both solutions can be the same value, which means that there is only one solution to the quadratic equation.





Quadratic Formula,
Quadratic Equation Solutions Examples



(1.1) 

Solve   x^{\tt{2}} + 5x \space {\text{–}} \space 6 = 0.

Solution   

This quadratic equation does happen to factor easily to   (x+6)(x \space {\text{–}} \space 1).

Giving the solutions   x = 1  and   x = {\text{-}}6.

But we can see how the quadratic equation solutions are obtained with the quadratic formula also.

Here:   a = 1 &nbsp,  b = 5  ,  c = -6.

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}5 \space \pm \space \sqrt{5^{2} \space – \space (4 \times 1 \times ({\text{-}}6))}}{2 \space \times \space 1}}}    {\scriptsize{=}} \space \space {\boldsymbol{\frac{{\text{-}}5 \space \pm \space \sqrt{25 \space + \space 24}}{2}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}5 \space + \space \sqrt{49}}{2}}}     ,     {\boldsymbol{\frac{{\text{-}}5 \space {\text{–}} \space \sqrt{49}}{2}}}

x = 1   or   x = {\text{-}}6




(1.2) 

Solve   4x^{\tt{2}} + 7x + 2 = 0.

Solution   

Here:   a = 4  ,  b = 7  ,  c = 2.

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}7 \space \pm \space \sqrt{7^{2} \space – \space (4 \times 4 \times 2)}}{2 \space \times \space 4}}}    {\scriptsize{=}} \space \space {\boldsymbol{\frac{{\text{-}}7 \space \pm \space \sqrt{49 \space {\text{–}} \space 32}}{8}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}7 \space + \space \sqrt{17}}{8}}}     or     {\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}7 \space {\text{–}} \space \sqrt{17}}{8}}}




(1.3) 

Solve   x^{\tt{2}} \space {\text{–}} \space 4x + 4 = 0.

Solution   

Here:   a = 1    b = -4  ,  c = 4.

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}({\text{-}}4) \space \pm \space \sqrt{({\text{-}}4)^{2} \space – \space (4 \times 1 \times 4)}}{2 \space \times \space 1}}}    {\scriptsize{=}} \space \space {\boldsymbol{\frac{4 \space \pm \space \sqrt{16 \space {\text{–}} \space 16}}{2}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{4 \space + \space 0}{2}}}     or     {\scriptsize{x = \space}} {\boldsymbol{\frac{4 \space {\text{–}} \space 0}{2}}}

There is only one solution to this quadratic equation solutions example, which is  x = 2.







Quadratic Equation Solutions,
No Real Solution


It is also possible that there can be no solution or solutions to a quadratic equation in Math.


We can consider the quadratic equation   x^{\tt{2}} + 4x + 6 = 0,
and apply the quadratic formula.

a = 1  ,  b = 4  ,  c = 6.

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}(4) \space \pm \space \sqrt{(4)^{2} \space – \space (4 \times 1 \times 6)}}{2 \space \times \space 1}}}    {\scriptsize{=}} \space \space {\boldsymbol{\frac{{\text{-}}4 \space \pm \space \sqrt{16 \space {\text{–}} \space 24}}{2}}}

    {\scriptsize{=}} \space \space {\boldsymbol{\frac{{\text{-}}4 \space \pm \space \sqrt{{\text{-}}8}}{2}}}



Here we have the square root of a negative number, which can be converted to a complex number.
The process of which is shown on the complex numbers page.

It’s the case that  \sqrt{{\text{-}}8}  can be written as the complex number  2\sqrt{2}i.

So we get:

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}4 \space + \space 2{\sqrt{2}}i}{2}}}     or     {\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}4 \space {\text{–}} \space 2{\sqrt{2}}i}{2}}}


Which can be simplified a bit further.

x = {\boldsymbol{{\text{-}}2 + {\sqrt{2}}i}}      or      x = {\boldsymbol{{\text{-}}2 – {\sqrt{2}}i}}



If we look at an image of the graph of   x^{\tt{2}} + 4x + 6.
We can see that it does not have any values where it touches the  x-axis.

Which is why there are no real solutions to the equation, but instead complex solutions.

Graph of a quadratic where there are no quadratic equation solutions that are real.





  1. Home
  2.  ›
  3. Algebra 1
  4. › Quadratic Solutions




Return to TOP of page