# How do you solve using the completing the square method 2x^2+3x=20?

Jun 11, 2016

$x = 2 \frac{1}{2} \text{ or } x = - 4$

#### Explanation:

Completing the square is based on the fact that when a binomial is squared, there is a specific relationship between the coefficients of the 2nd and 3rd terms."

${\left(x - 5\right)}^{2} = {x}^{2} - 10 x + 25$
Note that (-10) divided by 2 and then squared gives 25.

We have $2 {x}^{2} + 3 x = 20 \text{ divide by 2 first to get } {x}^{2}$

${x}^{2} + \frac{3}{2} x \text{ " = 10" }$ (3/2)÷ 2 = (3/4)

Add ${\left(\frac{3}{4}\right)}^{2}$ to both sides
${x}^{2} + \frac{3}{2} x + \textcolor{red}{{\left(\frac{3}{4}\right)}^{2}} = 10 + \textcolor{red}{{\left(\frac{3}{4}\right)}^{2}}$
The left side can now be written as ${\text{(binomial)}}^{2}$

${\left(x + \frac{3}{4}\right)}^{2} \text{ } = \frac{169}{16}$

$x + \frac{3}{4} = \pm \left(\frac{13}{4}\right) \text{ find the square root of both sides}$

$x = \frac{13}{4} - \frac{3}{4} \text{ or } x = - \frac{13}{4} - \frac{3}{4}$

$x = \frac{10}{4} \text{ or } x = - \frac{16}{4}$

$x = 2 \frac{1}{2} \text{ or } x = - 4$