0=-2p^2+22p-36

Solution for 0=-2p^2+22p-36 equation:

0=-2p^2+22p-36

We move all terms to the left:

0-(-2p^2+22p-36)=0

We add all the numbers together, and all the variables

-(-2p^2+22p-36)=0

We get rid of parentheses

2p^2-22p+36=0

a = 2; b = -22; c = +36;
Δ = b2-4ac
Δ = -222-4·2·36
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-14}{2*2}=\frac{8}{4}
=2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+14}{2*2}=\frac{36}{4}
=9
$