How do you factor $2x^2+5x=8$ ?

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Answer 1 · 2018-06-07T21:35:19

$x=(-5+-sqrt(89))/(4)$

$2x^2+5x=8$

$2x^2+5x-8=0$

At this point you cannot factor directly or by grouping so you must either complete the square or use the quadratic formula.

$ax^2+bx+c$

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-5+-sqrt(5^2-4*2*-8))/(2*2)$

$x=(-5+-sqrt(89))/(4)$

How do you factor 2x^2+5x=8 2x^2+5x=8?