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2q^2-11q=-7
We move all terms to the left:
2q^2-11q-(-7)=0
We add all the numbers together, and all the variables
2q^2-11q+7=0
a = 2; b = -11; c = +7;
Δ = b2-4ac
Δ = -112-4·2·7
Δ = 65
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{65}}{2*2}=\frac{11-\sqrt{65}}{4} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{65}}{2*2}=\frac{11+\sqrt{65}}{4} $
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