2x2+40x=88

Solution for 2x2+40x=88 equation:

2x^2+40x=88

We move all terms to the left:

2x^2+40x-(88)=0

a = 2; b = 40; c = -88;
Δ = b2-4ac
Δ = 402-4·2·(-88)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-48}{2*2}=\frac{-88}{4}
=-22
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+48}{2*2}=\frac{8}{4}
=2
$