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4y^2+27y-40=0
a = 4; b = 27; c = -40;
Δ = b2-4ac
Δ = 272-4·4·(-40)
Δ = 1369
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{1369}=37$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-37}{2*4}=\frac{-64}{8} =-8 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+37}{2*4}=\frac{10}{8} =1+1/4 $
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