24-8q+q^2=24-16q+3q^2

Solution for 24-8q+q^2=24-16q+3q^2 equation:

24-8q+q^2=24-16q+3q^2

We move all terms to the left:

24-8q+q^2-(24-16q+3q^2)=0

We get rid of parentheses

q^2-3q^2+16q-8q-24+24=0

We add all the numbers together, and all the variables

-2q^2+8q=0

a = -2; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·(-2)·0
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{-(8)-8}{2*-2}=\frac{-16}{-4}
=+4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-2}=\frac{0}{-4}
=0
$