P(x)=0.2x^2+20x-4

Solution for P(x)=0.2x^2+20x-4 equation:

(P)=0.2P^2+20P-4

We move all terms to the left:

(P)-(0.2P^2+20P-4)=0

We get rid of parentheses

-0.2P^2+P-20P+4=0

We add all the numbers together, and all the variables

-0.2P^2-19P+4=0

a = -0.2; b = -19; c = +4;
Δ = b2-4ac
Δ = -192-4·(-0.2)·4
Δ = 364.2
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{-(-19)-\sqrt{364.2}}{2*-0.2}=\frac{19-\sqrt{364.2}}{-0.4}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{364.2}}{2*-0.2}=\frac{19+\sqrt{364.2}}{-0.4}
$