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(3x-4)(3x-12)=0
We multiply parentheses ..
(+9x^2-36x-12x+48)=0
We get rid of parentheses
9x^2-36x-12x+48=0
We add all the numbers together, and all the variables
9x^2-48x+48=0
a = 9; b = -48; c = +48;
Δ = b2-4ac
Δ = -482-4·9·48
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-24}{2*9}=\frac{24}{18} =1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+24}{2*9}=\frac{72}{18} =4 $
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