P=0.5x^2+25x-150

Solution for P=0.5x^2+25x-150 equation:

=0.5P^2+25P-150

We move all terms to the left:

-(0.5P^2+25P-150)=0

We get rid of parentheses

-0.5P^2-25P+150=0

a = -0.5; b = -25; c = +150;
Δ = b2-4ac
Δ = -252-4·(-0.5)·150
Δ = 925
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{925}=\sqrt{25*37}=\sqrt{25}*\sqrt{37}=5\sqrt{37}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5\sqrt{37}}{2*-0.5}=\frac{25-5\sqrt{37}}{-1}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5\sqrt{37}}{2*-0.5}=\frac{25+5\sqrt{37}}{-1}
$