P=-t^2+17t+140

Solution for P=-t^2+17t+140 equation:

=-P^2+17P+140

We move all terms to the left:

-(-P^2+17P+140)=0

We get rid of parentheses

P^2-17P-140=0

a = 1; b = -17; c = -140;
Δ = b2-4ac
Δ = -172-4·1·(-140)
Δ = 849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-\sqrt{849}}{2*1}=\frac{17-\sqrt{849}}{2}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{849}}{2*1}=\frac{17+\sqrt{849}}{2}
$