12x^2+16x-528=0

Solution for 12x^2+16x-528=0 equation:

12x^2+16x-528=0

a = 12; b = 16; c = -528;
Δ = b2-4ac
Δ = 162-4·12·(-528)
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-160}{2*12}=\frac{-176}{24}
=-7+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+160}{2*12}=\frac{144}{24}
=6
$