(2-q)(2-q)+q^2=10

Solution for (2-q)(2-q)+q^2=10 equation:

(2-q)(2-q)+q^2=10

We move all terms to the left:

(2-q)(2-q)+q^2-(10)=0

We add all the numbers together, and all the variables

q^2+(-1q+2)(-1q+2)-10=0

We multiply parentheses ..

q^2+(+q^2-2q-2q+4)-10=0

We get rid of parentheses

q^2+q^2-2q-2q+4-10=0

We add all the numbers together, and all the variables

2q^2-4q-6=0

a = 2; b = -4; c = -6;
Δ = b2-4ac
Δ = -42-4·2·(-6)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*2}=\frac{-4}{4}
=-1
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*2}=\frac{12}{4}
=3
$