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(3x+1)+(1/2x+3)=18
We move all terms to the left:
(3x+1)+(1/2x+3)-(18)=0
Domain of the equation: 2x+3)!=0We get rid of parentheses
x∈R
3x+1/2x+1+3-18=0
We multiply all the terms by the denominator
3x*2x+1*2x+3*2x-18*2x+1=0
Wy multiply elements
6x^2+2x+6x-36x+1=0
We add all the numbers together, and all the variables
6x^2-28x+1=0
a = 6; b = -28; c = +1;
Δ = b2-4ac
Δ = -282-4·6·1
Δ = 760
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{760}=\sqrt{4*190}=\sqrt{4}*\sqrt{190}=2\sqrt{190}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2\sqrt{190}}{2*6}=\frac{28-2\sqrt{190}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2\sqrt{190}}{2*6}=\frac{28+2\sqrt{190}}{12} $
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