15=-2t^2+20t+15

Solution for 15=-2t^2+20t+15 equation:

15=-2t^2+20t+15

We move all terms to the left:

15-(-2t^2+20t+15)=0

We get rid of parentheses

2t^2-20t-15+15=0

We add all the numbers together, and all the variables

2t^2-20t=0

a = 2; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·2·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*2}=\frac{0}{4}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*2}=\frac{40}{4}
=10
$