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(x-4)(3x+6)+(x-4)(12x+48)=0
We multiply parentheses ..
(+3x^2+6x-12x-24)+(x-4)(12x+48)=0
We get rid of parentheses
3x^2+6x-12x+(x-4)(12x+48)-24=0
We multiply parentheses ..
3x^2+(+12x^2+48x-48x-192)+6x-12x-24=0
We add all the numbers together, and all the variables
3x^2+(+12x^2+48x-48x-192)-6x-24=0
We get rid of parentheses
3x^2+12x^2+48x-48x-6x-192-24=0
We add all the numbers together, and all the variables
15x^2-6x-216=0
a = 15; b = -6; c = -216;
Δ = b2-4ac
Δ = -62-4·15·(-216)
Δ = 12996
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{12996}=114$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-114}{2*15}=\frac{-108}{30} =-3+3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+114}{2*15}=\frac{120}{30} =4 $
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