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4y^2-15y+14=0
a = 4; b = -15; c = +14;
Δ = b2-4ac
Δ = -152-4·4·14
Δ = 1
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{1}=1$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-1}{2*4}=\frac{14}{8} =1+3/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+1}{2*4}=\frac{16}{8} =2 $
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