3q=3q2-7

Solution for 3q=3q2-7 equation:

3q=3q^2-7

We move all terms to the left:

3q-(3q^2-7)=0

We get rid of parentheses

-3q^2+3q+7=0

a = -3; b = 3; c = +7;
Δ = b2-4ac
Δ = 32-4·(-3)·7
Δ = 93
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{93}}{2*-3}=\frac{-3-\sqrt{93}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{93}}{2*-3}=\frac{-3+\sqrt{93}}{-6}
$