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(2y-3)(3y-1)=8y
We move all terms to the left:
(2y-3)(3y-1)-(8y)=0
We add all the numbers together, and all the variables
-8y+(2y-3)(3y-1)=0
We multiply parentheses ..
(+6y^2-2y-9y+3)-8y=0
We get rid of parentheses
6y^2-2y-9y-8y+3=0
We add all the numbers together, and all the variables
6y^2-19y+3=0
a = 6; b = -19; c = +3;
Δ = b2-4ac
Δ = -192-4·6·3
Δ = 289
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{289}=17$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-17}{2*6}=\frac{2}{12} =1/6 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+17}{2*6}=\frac{36}{12} =3 $
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