P(x)=3x^2-7x+7,Q(x)=x-2

Solution for P(x)=3x^2-7x+7,Q(x)=x-2 equation:

(P)=3P^2-7P+7.(P)=P-2

We move all terms to the left:

(P)-(3P^2-7P+7.(P))=0

We get rid of parentheses

-3P^2+P+7P-7.P=0

We add all the numbers together, and all the variables

-3P^2+P=0

a = -3; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-3)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-3}=\frac{-2}{-6}
=1/3
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-3}=\frac{0}{-6}
=0
$