# How do you solve 2x^2+4x-5=0?

Mar 25, 2016

Use the quadratic formula to find:

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

#### Explanation:

$2 {x}^{2} + 4 x - 5$ is in the form $a {x}^{2} + b x + c$ with $a = 2$, $b = 4$ and $c = - 5$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {4}^{2} - \left(4 \cdot 2 \cdot \left(- 5\right)\right) = 16 + 40 = 56 = {2}^{2} \cdot 14$

This is positive, but not a perfect square. So our quadratic equation has Real irrational roots.

We can find the roots using the quadratic formula:

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

$= \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 4 \pm \sqrt{56}}{2 \cdot 2}$

$= \frac{- 4 \pm 2 \sqrt{14}}{4}$

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

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Alternative Method

Multiply through by $2$ first to make the leading term a perfect square, complete the squre, then factorise using the difference of square identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(2 x + 2\right)$ and $b = \sqrt{14}$ as follows:

$0 = 4 {x}^{2} + 8 x - 10$

$= {\left(2 x + 2\right)}^{2} - 4 - 10$

$= {\left(2 x + 2\right)}^{2} - {\left(\sqrt{14}\right)}^{2}$

$= \left(\left(2 x + 2\right) - \sqrt{14}\right) \left(\left(2 x + 2\right) + \sqrt{14}\right)$

$= \left(2 x + 2 - \sqrt{14}\right) \left(2 x + 2 + \sqrt{14}\right)$

$= \left(2 \left(x + 1 - \frac{\sqrt{14}}{2}\right)\right) \left(2 \left(x + 1 + \frac{\sqrt{14}}{2}\right)\right)$

$= 4 \left(x + 1 - \frac{\sqrt{14}}{2}\right) \left(x + 1 + \frac{\sqrt{14}}{2}\right)$

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