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(8+y)y=105
We move all terms to the left:
(8+y)y-(105)=0
We add all the numbers together, and all the variables
(y+8)y-105=0
We multiply parentheses
y^2+8y-105=0
a = 1; b = 8; c = -105;
Δ = b2-4ac
Δ = 82-4·1·(-105)
Δ = 484
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{484}=22$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-22}{2*1}=\frac{-30}{2} =-15 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+22}{2*1}=\frac{14}{2} =7 $
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