How do you solve $4-3x=2x^2$ using the quadratic formula?

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Answer 1 · 2017-04-28T10:37:04

We first rearrange into a proper quadratic =0 equation:

$2x^2+3x-4=0$

$x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)$

$x_(1,2)=(-3+-sqrt(3^2-4*2*(-4)))/(2*2)$

$x_(1,2)=(-3+-sqrt(9+32))/4=-3/4+-1/4sqrt41$

graph{2x^2+3x-4 [-10.42, 9.58, -6.64, 3.36]}

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