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1/4x+3=3/2x-2
We move all terms to the left:
1/4x+3-(3/2x-2)=0
Domain of the equation: 4x!=0
x!=0/4
x!=0
x∈R
Domain of the equation: 2x-2)!=0We get rid of parentheses
x∈R
1/4x-3/2x+2+3=0
We calculate fractions
2x/8x^2+(-12x)/8x^2+2+3=0
We add all the numbers together, and all the variables
2x/8x^2+(-12x)/8x^2+5=0
We multiply all the terms by the denominator
2x+(-12x)+5*8x^2=0
Wy multiply elements
40x^2+2x+(-12x)=0
We get rid of parentheses
40x^2+2x-12x=0
We add all the numbers together, and all the variables
40x^2-10x=0
a = 40; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·40·0
Δ = 100
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*40}=\frac{0}{80} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*40}=\frac{20}{80} =1/4 $
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