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1/2x-4/5x=40
We move all terms to the left:
1/2x-4/5x-(40)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: 5x!=0We calculate fractions
x!=0/5
x!=0
x∈R
5x/10x^2+(-8x)/10x^2-40=0
We multiply all the terms by the denominator
5x+(-8x)-40*10x^2=0
Wy multiply elements
-400x^2+5x+(-8x)=0
We get rid of parentheses
-400x^2+5x-8x=0
We add all the numbers together, and all the variables
-400x^2-3x=0
a = -400; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-400)·0
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-400}=\frac{0}{-800} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-400}=\frac{6}{-800} =-3/400 $
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