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4y^2-4y=48
We move all terms to the left:
4y^2-4y-(48)=0
a = 4; b = -4; c = -48;
Δ = b2-4ac
Δ = -42-4·4·(-48)
Δ = 784
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{784}=28$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-28}{2*4}=\frac{-24}{8} =-3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+28}{2*4}=\frac{32}{8} =4 $
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