3q^2+4q-1=12

Solution for 3q^2+4q-1=12 equation:

3q^2+4q-1=12

We move all terms to the left:

3q^2+4q-1-(12)=0

We add all the numbers together, and all the variables

3q^2+4q-13=0

a = 3; b = 4; c = -13;
Δ = b2-4ac
Δ = 42-4·3·(-13)
Δ = 172
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:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{43}}{2*3}=\frac{-4-2\sqrt{43}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{43}}{2*3}=\frac{-4+2\sqrt{43}}{6}
$