1/5x-16=9/10x+12

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Solution for 1/5x-16=9/10x+12 equation:



1/5x-16=9/10x+12
We move all terms to the left:
1/5x-16-(9/10x+12)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 10x+12)!=0
x∈R
We get rid of parentheses
1/5x-9/10x-12-16=0
We calculate fractions
10x/50x^2+(-45x)/50x^2-12-16=0
We add all the numbers together, and all the variables
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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