P(x)=2x^2-1Q(x)=3x^2+x

Solution for P(x)=2x^2-1Q(x)=3x^2+x equation:

(P)=2P^2-1(P)=3P^2+P

We move all terms to the left:

(P)-(2P^2-1(P))=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2P^2+2P=0

a = -2; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-2)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-2}=\frac{-4}{-4}
=1
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-2}=\frac{0}{-4}
=0
$