15=20t+5t^2

Solution for 15=20t+5t^2 equation:

15=20t+5t^2

We move all terms to the left:

15-(20t+5t^2)=0

We get rid of parentheses

-5t^2-20t+15=0

a = -5; b = -20; c = +15;
Δ = b2-4ac
Δ = -202-4·(-5)·15
Δ = 700
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:

$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{7}}{2*-5}=\frac{20-10\sqrt{7}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{7}}{2*-5}=\frac{20+10\sqrt{7}}{-10}
$