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4y^2-52y+160=0
a = 4; b = -52; c = +160;
Δ = b2-4ac
Δ = -522-4·4·160
Δ = 144
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{144}=12$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-12}{2*4}=\frac{40}{8} =5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+12}{2*4}=\frac{64}{8} =8 $
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