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

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Answer 1 · 2016-06-04T20:00:12

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

We have

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

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

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

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

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

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

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

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

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

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

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

How do you solve 2x2+3x=202x^2+3x=20 by completing the square?