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

Solution for 30(t)=-t^2-2t+30 equation:

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

We move all terms to the left:

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

We get rid of parentheses

t^2+2t+30t-30=0

We add all the numbers together, and all the variables

t^2+32t-30=0

a = 1; b = 32; c = -30;
Δ = b2-4ac
Δ = 322-4·1·(-30)
Δ = 1144
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:

$\sqrt{\Delta}=\sqrt{1144}=\sqrt{4*286}=\sqrt{4}*\sqrt{286}=2\sqrt{286}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-2\sqrt{286}}{2*1}=\frac{-32-2\sqrt{286}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+2\sqrt{286}}{2*1}=\frac{-32+2\sqrt{286}}{2}
$