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

Jun 4, 2016

Factor out the two and cancel it from both sides to get
${x}^{2} + \frac{3}{2} x = 10$

#### Explanation:

We have

$2 {x}^{2} + 3 x = 20$

We factor out
$2 \left({x}^{2} + \frac{3}{2} x\right) = 20$

${x}^{2} + \frac{3}{2} x = 10$

Divide split the linear term up so we have $b = 2 \cdot \left(\frac{b}{2}\right)$
${x}^{2} + 2 \cdot \left(\frac{3}{4}\right) x = 10$

${x}^{2} + 2 \cdot \left(\frac{3}{4}\right) x + {3}^{2} / {4}^{2} = 10 + {3}^{2} / {4}^{2}$

${x}^{2} + 2 \cdot \left(\frac{3}{4}\right) x + {3}^{2} / {4}^{2} = 10 + \frac{9}{16}$

${x}^{2} + 2 \cdot \left(\frac{3}{4}\right) x + {3}^{2} / {4}^{2} = \frac{160}{16} + \frac{9}{16}$

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

$\left(x + \frac{3}{4}\right) = \pm \frac{13}{4}$

$x + \frac{3}{4} - \frac{3}{4} = \pm \frac{13}{4} - \frac{3}{4}$

$x = - \frac{3}{4} \pm \frac{13}{4}$