0.75t^2+12t-12=0

Solution for 0.75t^2+12t-12=0 equation:

0.75t^2+12t-12=0

a = 0.75; b = 12; c = -12;
Δ = b2-4ac
Δ = 122-4·0.75·(-12)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{5}}{2*0.75}=\frac{-12-6\sqrt{5}}{1.5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{5}}{2*0.75}=\frac{-12+6\sqrt{5}}{1.5}
$