# How do you solve 2x^2+3x-5=0 by completing the square?

Dec 4, 2016

$x = 1 \text{ }$ or $\text{ } x = - \frac{5}{2}$

#### Explanation:

The difference of squares identity can be written:

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

We use this later with $a = \left(4 x + 3\right)$ and $b = 7$

$\textcolor{w h i t e}{}$
Given:

$2 {x}^{2} + 3 x - 5 = 0$

To delay having to work with fractions, first multiply this by $8$.

Why $8$?

We want to multiply by $2$ to make the leading term into a perfect square, namely $4 {x}^{2} = {\left(2 x\right)}^{2}$. Then the middle term becomes $6 x = 3 \left(2 x\right)$. We want the coefficient $3$ of $\left(2 x\right)$ to be even too, so multiply by another factor of ${2}^{2} = 4$ to keep the leading term a perfect square.

Then we find:

$0 = 8 \left(2 {x}^{2} + 3 x - 5\right)$

$\textcolor{w h i t e}{0} = 16 {x}^{2} + 24 x - 40$

$\textcolor{w h i t e}{0} = {\left(4 x\right)}^{2} + 2 \left(3\right) \left(4 x\right) + 9 - 49$

$\textcolor{w h i t e}{0} = {\left(4 x + 3\right)}^{2} - {7}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(4 x + 3\right) - 7\right) \left(\left(4 x + 3\right) + 7\right)$

$\textcolor{w h i t e}{0} = \left(4 x - 4\right) \left(4 x + 10\right)$

$\textcolor{w h i t e}{0} = 4 \left(x - 1\right) 2 \left(2 x + 5\right)$

Hence:

$x = 1 \text{ }$ or $\text{ } x = - \frac{5}{2}$