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2/3x+9=1x+5
We move all terms to the left:
2/3x+9-(1x+5)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
2/3x-(x+5)+9=0
We get rid of parentheses
2/3x-x-5+9=0
We multiply all the terms by the denominator
-x*3x-5*3x+9*3x+2=0
Wy multiply elements
-3x^2-15x+27x+2=0
We add all the numbers together, and all the variables
-3x^2+12x+2=0
a = -3; b = 12; c = +2;
Δ = b2-4ac
Δ = 122-4·(-3)·2
Δ = 168
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{42}}{2*-3}=\frac{-12-2\sqrt{42}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{42}}{2*-3}=\frac{-12+2\sqrt{42}}{-6} $
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