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10-2q+q^2=130
We move all terms to the left:
10-2q+q^2-(130)=0
We add all the numbers together, and all the variables
q^2-2q-120=0
a = 1; b = -2; c = -120;
Δ = b2-4ac
Δ = -22-4·1·(-120)
Δ = 484
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{484}=22$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*1}=\frac{-20}{2} =-10 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*1}=\frac{24}{2} =12 $
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