32x^2+8x-15=0

Solution for 32x^2+8x-15=0 equation:

32x^2+8x-15=0

a = 32; b = 8; c = -15;
Δ = b2-4ac
Δ = 82-4·32·(-15)
Δ = 1984
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:

$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{31}}{2*32}=\frac{-8-8\sqrt{31}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{31}}{2*32}=\frac{-8+8\sqrt{31}}{64}
$