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8/3x+12/3x=x
We move all terms to the left:
8/3x+12/3x-(x)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
-1x+8/3x+12/3x=0
We multiply all the terms by the denominator
-1x*3x+8+12=0
We add all the numbers together, and all the variables
-1x*3x+20=0
Wy multiply elements
-3x^2+20=0
a = -3; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-3)·20
Δ = 240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*-3}=\frac{0-4\sqrt{15}}{-6} =-\frac{4\sqrt{15}}{-6} =-\frac{2\sqrt{15}}{-3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*-3}=\frac{0+4\sqrt{15}}{-6} =\frac{4\sqrt{15}}{-6} =\frac{2\sqrt{15}}{-3} $
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