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3q^2-2q=2q^2-4q+15
We move all terms to the left:
3q^2-2q-(2q^2-4q+15)=0
We get rid of parentheses
3q^2-2q^2-2q+4q-15=0
We add all the numbers together, and all the variables
q^2+2q-15=0
a = 1; b = 2; c = -15;
Δ = b2-4ac
Δ = 22-4·1·(-15)
Δ = 64
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{64}=8$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*1}=\frac{-10}{2} =-5 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*1}=\frac{6}{2} =3 $
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