3t^2-20t-24=0

Solution for 3t^2-20t-24=0 equation:

3t^2-20t-24=0

a = 3; b = -20; c = -24;
Δ = b2-4ac
Δ = -202-4·3·(-24)
Δ = 688
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:

$\sqrt{\Delta}=\sqrt{688}=\sqrt{16*43}=\sqrt{16}*\sqrt{43}=4\sqrt{43}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{43}}{2*3}=\frac{20-4\sqrt{43}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{43}}{2*3}=\frac{20+4\sqrt{43}}{6}
$