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2/5x-8=2-+3/4x
We move all terms to the left:
2/5x-8-(2-+3/4x)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 4x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
2/5x-(+3/4x)-8=0
We get rid of parentheses
2/5x-3/4x-8=0
We calculate fractions
8x/20x^2+(-15x)/20x^2-8=0
We multiply all the terms by the denominator
8x+(-15x)-8*20x^2=0
Wy multiply elements
-160x^2+8x+(-15x)=0
We get rid of parentheses
-160x^2+8x-15x=0
We add all the numbers together, and all the variables
-160x^2-7x=0
a = -160; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·(-160)·0
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*-160}=\frac{0}{-320} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*-160}=\frac{14}{-320} =-7/160 $
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