16x^2+40x-56=0

Solution for 16x^2+40x-56=0 equation:

16x^2+40x-56=0

a = 16; b = 40; c = -56;
Δ = b2-4ac
Δ = 402-4·16·(-56)
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*16}=\frac{-112}{32}
=-3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+72}{2*16}=\frac{32}{32}
=1
$