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1/5x-9/10x=25-53
We move all terms to the left:
1/5x-9/10x-(25-53)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 10x!=0We add all the numbers together, and all the variables
x!=0/10
x!=0
x∈R
1/5x-9/10x-(-28)=0
We add all the numbers together, and all the variables
1/5x-9/10x+28=0
We calculate fractions
10x/50x^2+(-45x)/50x^2+28=0
We multiply all the terms by the denominator
10x+(-45x)+28*50x^2=0
Wy multiply elements
1400x^2+10x+(-45x)=0
We get rid of parentheses
1400x^2+10x-45x=0
We add all the numbers together, and all the variables
1400x^2-35x=0
a = 1400; b = -35; c = 0;
Δ = b2-4ac
Δ = -352-4·1400·0
Δ = 1225
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{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-35}{2*1400}=\frac{0}{2800} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+35}{2*1400}=\frac{70}{2800} =1/40 $
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