0=t^2+5t=36

Solution for 0=t^2+5t=36 equation:

0=t^2+5t=36

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(t^2+5t)=0

We get rid of parentheses

-t^2-5t=0

We add all the numbers together, and all the variables

-1t^2-5t=0

a = -1; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-1)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-1}=\frac{0}{-2}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-1}=\frac{10}{-2}
=-5
$