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y^2-13y-68=0
a = 1; b = -13; c = -68;
Δ = b2-4ac
Δ = -132-4·1·(-68)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{441}=21$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-21}{2*1}=\frac{-8}{2} =-4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+21}{2*1}=\frac{34}{2} =17 $
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