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5y^2-24y+16=0
a = 5; b = -24; c = +16;
Δ = b2-4ac
Δ = -242-4·5·16
Δ = 256
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{256}=16$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-16}{2*5}=\frac{8}{10} =4/5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+16}{2*5}=\frac{40}{10} =4 $
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