32x^2+32x-33=0

Solution for 32x^2+32x-33=0 equation:

32x^2+32x-33=0

a = 32; b = 32; c = -33;
Δ = b2-4ac
Δ = 322-4·32·(-33)
Δ = 5248
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:

$\sqrt{\Delta}=\sqrt{5248}=\sqrt{64*82}=\sqrt{64}*\sqrt{82}=8\sqrt{82}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{82}}{2*32}=\frac{-32-8\sqrt{82}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{82}}{2*32}=\frac{-32+8\sqrt{82}}{64}
$