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1/2x+3=x+4
We move all terms to the left:
1/2x+3-(x+4)=0
Domain of the equation: 2x!=0We get rid of parentheses
x!=0/2
x!=0
x∈R
1/2x-x-4+3=0
We multiply all the terms by the denominator
-x*2x-4*2x+3*2x+1=0
Wy multiply elements
-2x^2-8x+6x+1=0
We add all the numbers together, and all the variables
-2x^2-2x+1=0
a = -2; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-2)·1
Δ = 12
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{3}}{2*-2}=\frac{2-2\sqrt{3}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{3}}{2*-2}=\frac{2+2\sqrt{3}}{-4} $
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