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y^2+14y-240=0
a = 1; b = 14; c = -240;
Δ = b2-4ac
Δ = 142-4·1·(-240)
Δ = 1156
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{1156}=34$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-34}{2*1}=\frac{-48}{2} =-24 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+34}{2*1}=\frac{20}{2} =10 $
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