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4y^2+12y-16=0
a = 4; b = 12; c = -16;
Δ = b2-4ac
Δ = 122-4·4·(-16)
Δ = 400
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{400}=20$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-20}{2*4}=\frac{-32}{8} =-4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+20}{2*4}=\frac{8}{8} =1 $
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