13(t)=5t^2+20t

Solution for 13(t)=5t^2+20t equation:

13(t)=5t^2+20t

We move all terms to the left:

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

We get rid of parentheses

-5t^2+13t-20t=0

We add all the numbers together, and all the variables

-5t^2-7t=0

a = -5; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·(-5)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*-5}=\frac{0}{-10}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*-5}=\frac{14}{-10}
=-1+2/5
$