t2+15t=-36

Solution for t2+15t=-36 equation:

t2+15t=-36

We move all terms to the left:

t2+15t-(-36)=0

We add all the numbers together, and all the variables

t^2+15t+36=0

a = 1; b = 15; c = +36;
Δ = b2-4ac
Δ = 152-4·1·36
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-9}{2*1}=\frac{-24}{2}
=-12
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+9}{2*1}=\frac{-6}{2}
=-3
$