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(2y-5)(9+y)=0
We add all the numbers together, and all the variables
(2y-5)(y+9)=0
We multiply parentheses ..
(+2y^2+18y-5y-45)=0
We get rid of parentheses
2y^2+18y-5y-45=0
We add all the numbers together, and all the variables
2y^2+13y-45=0
a = 2; b = 13; c = -45;
Δ = b2-4ac
Δ = 132-4·2·(-45)
Δ = 529
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{529}=23$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-23}{2*2}=\frac{-36}{4} =-9 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+23}{2*2}=\frac{10}{4} =2+1/2 $
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