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y^2+12y=220
We move all terms to the left:
y^2+12y-(220)=0
a = 1; b = 12; c = -220;
Δ = b2-4ac
Δ = 122-4·1·(-220)
Δ = 1024
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{1024}=32$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-32}{2*1}=\frac{-44}{2} =-22 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+32}{2*1}=\frac{20}{2} =10 $
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