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(y+3)(2y+3)-2y=18
We move all terms to the left:
(y+3)(2y+3)-2y-(18)=0
We add all the numbers together, and all the variables
-2y+(y+3)(2y+3)-18=0
We multiply parentheses ..
(+2y^2+3y+6y+9)-2y-18=0
We get rid of parentheses
2y^2+3y+6y-2y+9-18=0
We add all the numbers together, and all the variables
2y^2+7y-9=0
a = 2; b = 7; c = -9;
Δ = b2-4ac
Δ = 72-4·2·(-9)
Δ = 121
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{121}=11$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-11}{2*2}=\frac{-18}{4} =-4+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+11}{2*2}=\frac{4}{4} =1 $
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