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.

(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,

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.

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