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q^2-10q-96=0
a = 1; b = -10; c = -96;
Δ = b2-4ac
Δ = -102-4·1·(-96)
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$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*1}=\frac{-12}{2} =-6 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*1}=\frac{32}{2} =16 $
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