6t^2+6t-32=0

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

6t^2+6t-32=0

a = 6; b = 6; c = -32;
Δ = b2-4ac
Δ = 62-4·6·(-32)
Δ = 804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{201}}{2*6}=\frac{-6-2\sqrt{201}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{201}}{2*6}=\frac{-6+2\sqrt{201}}{12}
$