8q^2-4q=7q^2+7q+54

Solution for 8q^2-4q=7q^2+7q+54 equation:

8q^2-4q=7q^2+7q+54

We move all terms to the left:

8q^2-4q-(7q^2+7q+54)=0

We get rid of parentheses

8q^2-7q^2-4q-7q-54=0

We add all the numbers together, and all the variables

q^2-11q-54=0

a = 1; b = -11; c = -54;
Δ = b2-4ac
Δ = -112-4·1·(-54)
Δ = 337
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{-(-11)-\sqrt{337}}{2*1}=\frac{11-\sqrt{337}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{337}}{2*1}=\frac{11+\sqrt{337}}{2}
$