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

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

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

a = -16; b = -16; c = +12;
Δ = b2-4ac
Δ = -162-4·(-16)·12
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-32}{2*-16}=\frac{-16}{-32}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+32}{2*-16}=\frac{48}{-32}
=-1+1/2
$