24=-4t2+20t

Solution for 24=-4t2+20t equation:

24=-4t^2+20t

We move all terms to the left:

24-(-4t^2+20t)=0

We get rid of parentheses

4t^2-20t+24=0

a = 4; b = -20; c = +24;
Δ = b2-4ac
Δ = -202-4·4·24
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*4}=\frac{16}{8}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*4}=\frac{24}{8}
=3
$