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5/6x=5/8x+10
We move all terms to the left:
5/6x-(5/8x+10)=0
Domain of the equation: 6x!=0
x!=0/6
x!=0
x∈R
Domain of the equation: 8x+10)!=0We get rid of parentheses
x∈R
5/6x-5/8x-10=0
We calculate fractions
40x/48x^2+(-30x)/48x^2-10=0
We multiply all the terms by the denominator
40x+(-30x)-10*48x^2=0
Wy multiply elements
-480x^2+40x+(-30x)=0
We get rid of parentheses
-480x^2+40x-30x=0
We add all the numbers together, and all the variables
-480x^2+10x=0
a = -480; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-480)·0
Δ = 100
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{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-480}=\frac{-20}{-960} =1/48 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-480}=\frac{0}{-960} =0 $
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