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4/5x-3/4x=8
We move all terms to the left:
4/5x-3/4x-(8)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 4x!=0We calculate fractions
x!=0/4
x!=0
x∈R
16x/20x^2+(-15x)/20x^2-8=0
We multiply all the terms by the denominator
16x+(-15x)-8*20x^2=0
Wy multiply elements
-160x^2+16x+(-15x)=0
We get rid of parentheses
-160x^2+16x-15x=0
We add all the numbers together, and all the variables
-160x^2+x=0
a = -160; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-160)·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-160}=\frac{-2}{-320} =1/160 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-160}=\frac{0}{-320} =0 $
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