-16x^2+12x+40=0

Solution for -16x^2+12x+40=0 equation:

-16x^2+12x+40=0

a = -16; b = 12; c = +40;
Δ = b2-4ac
Δ = 122-4·(-16)·40
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-52}{2*-16}=\frac{-64}{-32}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+52}{2*-16}=\frac{40}{-32}
=-1+1/4
$