P(x)=(x+3)(x-5)

Solution for P(x)=(x+3)(x-5) equation:

(P)=(P+3)(P-5)

We move all terms to the left:

(P)-((P+3)(P-5))=0

We multiply parentheses ..

-((+P^2-5P+3P-15))+P=0


We calculate terms in parentheses: -((+P^2-5P+3P-15)), so:

(+P^2-5P+3P-15)

We get rid of parentheses

P^2-5P+3P-15

We add all the numbers together, and all the variables

P^2-2P-15

Back to the equation:

-(P^2-2P-15)


We add all the numbers together, and all the variables

P-(P^2-2P-15)=0

We get rid of parentheses

-P^2+P+2P+15=0

We add all the numbers together, and all the variables

-1P^2+3P+15=0

a = -1; b = 3; c = +15;
Δ = b2-4ac
Δ = 32-4·(-1)·15
Δ = 69
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{-(3)-\sqrt{69}}{2*-1}=\frac{-3-\sqrt{69}}{-2}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{69}}{2*-1}=\frac{-3+\sqrt{69}}{-2}
$