32x^2+12x-57=0

Solution for 32x^2+12x-57=0 equation:

32x^2+12x-57=0

a = 32; b = 12; c = -57;
Δ = b2-4ac
Δ = 122-4·32·(-57)
Δ = 7440
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:

$\sqrt{\Delta}=\sqrt{7440}=\sqrt{16*465}=\sqrt{16}*\sqrt{465}=4\sqrt{465}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{465}}{2*32}=\frac{-12-4\sqrt{465}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{465}}{2*32}=\frac{-12+4\sqrt{465}}{64}
$