12q^2-28=-34q

Solution for 12q^2-28=-34q equation:

12q^2-28=-34q

We move all terms to the left:

12q^2-28-(-34q)=0

We get rid of parentheses

12q^2+34q-28=0

a = 12; b = 34; c = -28;
Δ = b2-4ac
Δ = 342-4·12·(-28)
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-50}{2*12}=\frac{-84}{24}
=-3+1/2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+50}{2*12}=\frac{16}{24}
=2/3
$