3q^2-6Q=2Q^2-8Q+35

Solution for 3q^2-6Q=2Q^2-8Q+35 equation:

3q^2-6=2^2-8+35

We move all terms to the left:

3q^2-6-(2^2-8+35)=0

We add all the numbers together, and all the variables

3q^2-37=0

a = 3; b = 0; c = -37;
Δ = b2-4ac
Δ = 02-4·3·(-37)
Δ = 444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{444}=\sqrt{4*111}=\sqrt{4}*\sqrt{111}=2\sqrt{111}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{111}}{2*3}=\frac{0-2\sqrt{111}}{6}
=-\frac{2\sqrt{111}}{6}
=-\frac{\sqrt{111}}{3}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{111}}{2*3}=\frac{0+2\sqrt{111}}{6}
=\frac{2\sqrt{111}}{6}
=\frac{\sqrt{111}}{3}
$