6q^2-4q=5q^2-7q+28

Solution for 6q^2-4q=5q^2-7q+28 equation:

6q^2-4q=5q^2-7q+28

We move all terms to the left:

6q^2-4q-(5q^2-7q+28)=0

We get rid of parentheses

6q^2-5q^2-4q+7q-28=0

We add all the numbers together, and all the variables

q^2+3q-28=0

a = 1; b = 3; c = -28;
Δ = b2-4ac
Δ = 32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*1}=\frac{-14}{2}
=-7
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*1}=\frac{8}{2}
=4
$