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(y)(y+2)=168
We move all terms to the left:
(y)(y+2)-(168)=0
We multiply parentheses
y^2+2y-168=0
a = 1; b = 2; c = -168;
Δ = b2-4ac
Δ = 22-4·1·(-168)
Δ = 676
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{676}=26$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-26}{2*1}=\frac{-28}{2} =-14 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+26}{2*1}=\frac{24}{2} =12 $
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