0=-t^2+9t+52

Solution for 0=-t^2+9t+52 equation:

0=-t^2+9t+52

We move all terms to the left:

0-(-t^2+9t+52)=0

We add all the numbers together, and all the variables

-(-t^2+9t+52)=0

We get rid of parentheses

t^2-9t-52=0

a = 1; b = -9; c = -52;
Δ = b2-4ac
Δ = -92-4·1·(-52)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-17}{2*1}=\frac{-8}{2}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+17}{2*1}=\frac{26}{2}
=13
$