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1/2x+3=x-3
We move all terms to the left:
1/2x+3-(x-3)=0
Domain of the equation: 2x!=0We get rid of parentheses
x!=0/2
x!=0
x∈R
1/2x-x+3+3=0
We multiply all the terms by the denominator
-x*2x+3*2x+3*2x+1=0
Wy multiply elements
-2x^2+6x+6x+1=0
We add all the numbers together, and all the variables
-2x^2+12x+1=0
a = -2; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-2)·1
Δ = 152
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{38}}{2*-2}=\frac{-12-2\sqrt{38}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{38}}{2*-2}=\frac{-12+2\sqrt{38}}{-4} $
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