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1/2x-3=2x+3
We move all terms to the left:
1/2x-3-(2x+3)=0
Domain of the equation: 2x!=0We get rid of parentheses
x!=0/2
x!=0
x∈R
1/2x-2x-3-3=0
We multiply all the terms by the denominator
-2x*2x-3*2x-3*2x+1=0
Wy multiply elements
-4x^2-6x-6x+1=0
We add all the numbers together, and all the variables
-4x^2-12x+1=0
a = -4; b = -12; c = +1;
Δ = b2-4ac
Δ = -122-4·(-4)·1
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{10}}{2*-4}=\frac{12-4\sqrt{10}}{-8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{10}}{2*-4}=\frac{12+4\sqrt{10}}{-8} $
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