6x^2+32x-16=0

Solution for 6x^2+32x-16=0 equation:

6x^2+32x-16=0

a = 6; b = 32; c = -16;
Δ = b2-4ac
Δ = 322-4·6·(-16)
Δ = 1408
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:

$\sqrt{\Delta}=\sqrt{1408}=\sqrt{64*22}=\sqrt{64}*\sqrt{22}=8\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{22}}{2*6}=\frac{-32-8\sqrt{22}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{22}}{2*6}=\frac{-32+8\sqrt{22}}{12}
$