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(12x-3)(2x+4)=0
We multiply parentheses ..
(+24x^2+48x-6x-12)=0
We get rid of parentheses
24x^2+48x-6x-12=0
We add all the numbers together, and all the variables
24x^2+42x-12=0
a = 24; b = 42; c = -12;
Δ = b2-4ac
Δ = 422-4·24·(-12)
Δ = 2916
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{2916}=54$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-54}{2*24}=\frac{-96}{48} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+54}{2*24}=\frac{12}{48} =1/4 $
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