t^2+12t-12=0

Solution for t^2+12t-12=0 equation:

t^2+12t-12=0

a = 1; b = 12; c = -12;
Δ = b2-4ac
Δ = 122-4·1·(-12)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{3}}{2*1}=\frac{-12-8\sqrt{3}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{3}}{2*1}=\frac{-12+8\sqrt{3}}{2}
$