How do you solve $2x^2+3x=20$ 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
$x^2+3/2x=10$

We have

$2x^2+3x=20$

We factor out
$2(x^2+3/2x)=20$

$x^2+3/2x=10$

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

$x^2+2*(3/4)x+3^2/4^2=10+3^2/4^2$

$x^2+2*(3/4)x+3^2/4^2=10+9/16$

$x^2+2*(3/4)x+3^2/4^2=160/16+9/16$

$(x+3/4)^2=169/16$

$(x+3/4)=pm 13/4$

$x+3/4-3/4=pm 13/4-3/4$

$x=-3/4 pm 13/4$

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