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(3y-4)(7y-4)=0
We multiply parentheses ..
(+21y^2-12y-28y+16)=0
We get rid of parentheses
21y^2-12y-28y+16=0
We add all the numbers together, and all the variables
21y^2-40y+16=0
a = 21; b = -40; c = +16;
Δ = b2-4ac
Δ = -402-4·21·16
Δ = 256
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{256}=16$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-16}{2*21}=\frac{24}{42} =4/7 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+16}{2*21}=\frac{56}{42} =1+1/3 $
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