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2y(y-3)+5(y-3)=0
We multiply parentheses
2y^2-6y+5y-15=0
We add all the numbers together, and all the variables
2y^2-1y-15=0
a = 2; b = -1; c = -15;
Δ = b2-4ac
Δ = -12-4·2·(-15)
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*2}=\frac{-10}{4} =-2+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*2}=\frac{12}{4} =3 $
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