3g^2+32g-256=0

Solution for 3g^2+32g-256=0 equation:

3g^2+32g-256=0

a = 3; b = 32; c = -256;
Δ = b2-4ac
Δ = 322-4·3·(-256)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-64}{2*3}=\frac{-96}{6}
=-16
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+64}{2*3}=\frac{32}{6}
=5+1/3
$