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4y^2+36y+56=0
a = 4; b = 36; c = +56;
Δ = b2-4ac
Δ = 362-4·4·56
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-20}{2*4}=\frac{-56}{8} =-7 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+20}{2*4}=\frac{-16}{8} =-2 $
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