t^2+11t=-30

Solution for t^2+11t=-30 equation:

t^2+11t=-30

We move all terms to the left:

t^2+11t-(-30)=0

We add all the numbers together, and all the variables

t^2+11t+30=0

a = 1; b = 11; c = +30;
Δ = b2-4ac
Δ = 112-4·1·30
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*1}=\frac{-12}{2}
=-6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*1}=\frac{-10}{2}
=-5
$