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q^2+q-120=0
a = 1; b = 1; c = -120;
Δ = b2-4ac
Δ = 12-4·1·(-120)
Δ = 481
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{-(1)-\sqrt{481}}{2*1}=\frac{-1-\sqrt{481}}{2} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{481}}{2*1}=\frac{-1+\sqrt{481}}{2} $
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