12t^2-36t+8=0

Solution for 12t^2-36t+8=0 equation:

12t^2-36t+8=0

a = 12; b = -36; c = +8;
Δ = b2-4ac
Δ = -362-4·12·8
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{57}}{2*12}=\frac{36-4\sqrt{57}}{24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{57}}{2*12}=\frac{36+4\sqrt{57}}{24}
$