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2q^2-6=-3q
We move all terms to the left:
2q^2-6-(-3q)=0
We get rid of parentheses
2q^2+3q-6=0
a = 2; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·2·(-6)
Δ = 57
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{-(3)-\sqrt{57}}{2*2}=\frac{-3-\sqrt{57}}{4} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{57}}{2*2}=\frac{-3+\sqrt{57}}{4} $
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