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(6x+4)2x=4+x
We move all terms to the left:
(6x+4)2x-(4+x)=0
We add all the numbers together, and all the variables
(6x+4)2x-(x+4)=0
We multiply parentheses
12x^2+8x-(x+4)=0
We get rid of parentheses
12x^2+8x-x-4=0
We add all the numbers together, and all the variables
12x^2+7x-4=0
a = 12; b = 7; c = -4;
Δ = b2-4ac
Δ = 72-4·12·(-4)
Δ = 241
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{241}}{2*12}=\frac{-7-\sqrt{241}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{241}}{2*12}=\frac{-7+\sqrt{241}}{24} $
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