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(3y-5)(2y+9)=0
We multiply parentheses ..
(+6y^2+27y-10y-45)=0
We get rid of parentheses
6y^2+27y-10y-45=0
We add all the numbers together, and all the variables
6y^2+17y-45=0
a = 6; b = 17; c = -45;
Δ = b2-4ac
Δ = 172-4·6·(-45)
Δ = 1369
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{1369}=37$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-37}{2*6}=\frac{-54}{12} =-4+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+37}{2*6}=\frac{20}{12} =1+2/3 $
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