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6q^2-3q=5q^2-7q+32
We move all terms to the left:
6q^2-3q-(5q^2-7q+32)=0
We get rid of parentheses
6q^2-5q^2-3q+7q-32=0
We add all the numbers together, and all the variables
q^2+4q-32=0
a = 1; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·1·(-32)
Δ = 144
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{144}=12$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*1}=\frac{-16}{2} =-8 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*1}=\frac{8}{2} =4 $
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