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13-14y+y^2=0
a = 1; b = -14; c = +13;
Δ = b2-4ac
Δ = -142-4·1·13
Δ = 144
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{144}=12$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-12}{2*1}=\frac{2}{2} =1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+12}{2*1}=\frac{26}{2} =13 $
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