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2y^2+28y+48=0
a = 2; b = 28; c = +48;
Δ = b2-4ac
Δ = 282-4·2·48
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-20}{2*2}=\frac{-48}{4} =-12 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+20}{2*2}=\frac{-8}{4} =-2 $
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