-16x^2+40x+56=75

Solution for -16x^2+40x+56=75 equation:

-16x^2+40x+56=75

We move all terms to the left:

-16x^2+40x+56-(75)=0

We add all the numbers together, and all the variables

-16x^2+40x-19=0

a = -16; b = 40; c = -19;
Δ = b2-4ac
Δ = 402-4·(-16)·(-19)
Δ = 384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{6}}{2*-16}=\frac{-40-8\sqrt{6}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{6}}{2*-16}=\frac{-40+8\sqrt{6}}{-32}
$