# How do you solve 2x^2 - 4x - 5=0 using the quadratic formula?

Jan 18, 2017

$x = 1 \pm \frac{\sqrt{14}}{2}$

#### Explanation:

$2 {x}^{2} - 4 x - 5 = 0$

is of the form:

$a {x}^{2} + b x + c = 0$

with $a = 2$, $b = - 4$ and $c = - 5$.

This has roots given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(2\right) \left(- 5\right)}}{2 \left(2\right)}$

$\textcolor{w h i t e}{x} = \frac{4 \pm \sqrt{16 + 40}}{4}$

$\textcolor{w h i t e}{x} = \frac{4 \pm \sqrt{56}}{4}$

$\textcolor{w h i t e}{x} = \frac{4 \pm \sqrt{{2}^{2} \cdot 14}}{4}$

$\textcolor{w h i t e}{x} = \frac{4 \pm 2 \sqrt{14}}{4}$

$\textcolor{w h i t e}{x} = 1 \pm \frac{\sqrt{14}}{2}$

$\textcolor{w h i t e}{}$
Footnote

The quadratic formula is very useful, but is it just a "magical" formula to you, or do you know how to derive it?

Here's one way:

Given:

$a {x}^{2} + b x + c = 0$

We find:

$0 = \frac{1}{a} \left(a {x}^{2} + b x + c\right)$

$\textcolor{w h i t e}{0} = {x}^{2} + \frac{b}{a} x + \frac{c}{a}$

$\textcolor{w h i t e}{0} = {x}^{2} + 2 \frac{b}{2 a} x + {b}^{2} / {\left(2 a\right)}^{2} - {b}^{2} / {\left(2 a\right)}^{2} + \frac{c}{a}$

$\textcolor{w h i t e}{0} = {\left(x + \frac{b}{2 a}\right)}^{2} - \frac{{b}^{2} - 4 a c}{4 {a}^{2}}$

Add $\frac{{b}^{2} - 4 a c}{4 {a}^{2}}$ to both ends and transpose to get:

${\left(x + \frac{b}{2 a}\right)}^{2} = \frac{{b}^{2} - 4 a c}{4 {a}^{2}}$

Take the square root of both sides, allowing for both positive and negative square roots to find:

$x + \frac{b}{2 a} = \pm \sqrt{\frac{{b}^{2} - 4 a c}{4 {a}^{2}}} = \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$

Subtract $\frac{b}{2 a}$ from both ends to find:

$x = - \frac{b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$