32x^2+40x-12=0

Solution for 32x^2+40x-12=0 equation:

32x^2+40x-12=0

a = 32; b = 40; c = -12;
Δ = b2-4ac
Δ = 402-4·32·(-12)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-56}{2*32}=\frac{-96}{64}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+56}{2*32}=\frac{16}{64}
=1/4
$