32x^2-32x-14=0

Solution for 32x^2-32x-14=0 equation:

32x^2-32x-14=0

a = 32; b = -32; c = -14;
Δ = b2-4ac
Δ = -322-4·32·(-14)
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{11}}{2*32}=\frac{32-16\sqrt{11}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{11}}{2*32}=\frac{32+16\sqrt{11}}{64}
$