4q2+8q=4q+8

Solution for 4q2+8q=4q+8 equation:

4q^2+8q=4q+8

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4q^2+4q-8=0

a = 4; b = 4; c = -8;
Δ = b2-4ac
Δ = 42-4·4·(-8)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*4}=\frac{-16}{8}
=-2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*4}=\frac{8}{8}
=1
$