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4y^2-23y+29=0
a = 4; b = -23; c = +29;
Δ = b2-4ac
Δ = -232-4·4·29
Δ = 65
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{-(-23)-\sqrt{65}}{2*4}=\frac{23-\sqrt{65}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{65}}{2*4}=\frac{23+\sqrt{65}}{8} $
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