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(2x-12)(3x+4)=0
We multiply parentheses ..
(+6x^2+8x-36x-48)=0
We get rid of parentheses
6x^2+8x-36x-48=0
We add all the numbers together, and all the variables
6x^2-28x-48=0
a = 6; b = -28; c = -48;
Δ = b2-4ac
Δ = -282-4·6·(-48)
Δ = 1936
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{1936}=44$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-44}{2*6}=\frac{-16}{12} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+44}{2*6}=\frac{72}{12} =6 $
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