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9x-3/2x+3-4x+6/2x+3=5/1
We move all terms to the left:
9x-3/2x+3-4x+6/2x+3-(5/1)=0
Domain of the equation: 2x!=0We add all the numbers together, and all the variables
x!=0/2
x!=0
x∈R
9x-3/2x-4x+6/2x+3+3-0=0
We add all the numbers together, and all the variables
5x-3/2x+6/2x+6=0
We multiply all the terms by the denominator
5x*2x+6*2x-3+6=0
We add all the numbers together, and all the variables
5x*2x+6*2x+3=0
Wy multiply elements
10x^2+12x+3=0
a = 10; b = 12; c = +3;
Δ = b2-4ac
Δ = 122-4·10·3
Δ = 24
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{6}}{2*10}=\frac{-12-2\sqrt{6}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{6}}{2*10}=\frac{-12+2\sqrt{6}}{20} $
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