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4y^2-15y-11=0
a = 4; b = -15; c = -11;
Δ = b2-4ac
Δ = -152-4·4·(-11)
Δ = 401
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{401}}{2*4}=\frac{15-\sqrt{401}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{401}}{2*4}=\frac{15+\sqrt{401}}{8} $
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