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y^2+y-240=0
a = 1; b = 1; c = -240;
Δ = b2-4ac
Δ = 12-4·1·(-240)
Δ = 961
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{961}=31$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*1}=\frac{-32}{2} =-16 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*1}=\frac{30}{2} =15 $
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