50-8q+3q^2=50-4q+q^2

Solution for 50-8q+3q^2=50-4q+q^2 equation:

50-8q+3q^2=50-4q+q^2

We move all terms to the left:

50-8q+3q^2-(50-4q+q^2)=0

We get rid of parentheses

3q^2-q^2+4q-8q-50+50=0

We add all the numbers together, and all the variables

2q^2-4q=0

a = 2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·2·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*2}=\frac{0}{4}
=0
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*2}=\frac{8}{4}
=2
$