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40y^2+10y-15=0
a = 40; b = 10; c = -15;
Δ = b2-4ac
Δ = 102-4·40·(-15)
Δ = 2500
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{2500}=50$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-50}{2*40}=\frac{-60}{80} =-3/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+50}{2*40}=\frac{40}{80} =1/2 $
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