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(3y-4)(2y+1)=0
We multiply parentheses ..
(+6y^2+3y-8y-4)=0
We get rid of parentheses
6y^2+3y-8y-4=0
We add all the numbers together, and all the variables
6y^2-5y-4=0
a = 6; b = -5; c = -4;
Δ = b2-4ac
Δ = -52-4·6·(-4)
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*6}=\frac{-6}{12} =-1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*6}=\frac{16}{12} =1+1/3 $
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