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q^2-14q=51
We move all terms to the left:
q^2-14q-(51)=0
a = 1; b = -14; c = -51;
Δ = b2-4ac
Δ = -142-4·1·(-51)
Δ = 400
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{400}=20$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-20}{2*1}=\frac{-6}{2} =-3 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+20}{2*1}=\frac{34}{2} =17 $
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