6(t^2+t-20)=0

Solution for 6(t^2+t-20)=0 equation:

6(t^2+t-20)=0

We multiply parentheses

6t^2+6t-120=0

a = 6; b = 6; c = -120;
Δ = b2-4ac
Δ = 62-4·6·(-120)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-54}{2*6}=\frac{-60}{12}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+54}{2*6}=\frac{48}{12}
=4
$