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4y^2-25y+27=0
a = 4; b = -25; c = +27;
Δ = b2-4ac
Δ = -252-4·4·27
Δ = 193
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{-(-25)-\sqrt{193}}{2*4}=\frac{25-\sqrt{193}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{193}}{2*4}=\frac{25+\sqrt{193}}{8} $
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