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4y^2+8y-96=0
a = 4; b = 8; c = -96;
Δ = b2-4ac
Δ = 82-4·4·(-96)
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-40}{2*4}=\frac{-48}{8} =-6 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+40}{2*4}=\frac{32}{8} =4 $
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