8x^2+40x=112

Solution for 8x^2+40x=112 equation:

8x^2+40x=112

We move all terms to the left:

8x^2+40x-(112)=0

a = 8; b = 40; c = -112;
Δ = b2-4ac
Δ = 402-4·8·(-112)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-72}{2*8}=\frac{-112}{16}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+72}{2*8}=\frac{32}{16}
=2
$