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15y=12y(y-3)
We move all terms to the left:
15y-(12y(y-3))=0
We calculate terms in parentheses: -(12y(y-3)), so:We get rid of parentheses
12y(y-3)
We multiply parentheses
12y^2-36y
Back to the equation:
-(12y^2-36y)
-12y^2+15y+36y=0
We add all the numbers together, and all the variables
-12y^2+51y=0
a = -12; b = 51; c = 0;
Δ = b2-4ac
Δ = 512-4·(-12)·0
Δ = 2601
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{2601}=51$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-51}{2*-12}=\frac{-102}{-24} =4+1/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+51}{2*-12}=\frac{0}{-24} =0 $
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