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4y^2+24y-64=0
a = 4; b = 24; c = -64;
Δ = b2-4ac
Δ = 242-4·4·(-64)
Δ = 1600
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{1600}=40$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-40}{2*4}=\frac{-64}{8} =-8 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+40}{2*4}=\frac{16}{8} =2 $
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