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y^2-19y+60=0
a = 1; b = -19; c = +60;
Δ = b2-4ac
Δ = -192-4·1·60
Δ = 121
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{121}=11$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*1}=\frac{8}{2} =4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*1}=\frac{30}{2} =15 $
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