2/5x+10=1/4x+40

Solution for 2/5x+10=1/4x+40 equation:

2/5x+10=1/4x+40

We move all terms to the left:

2/5x+10-(1/4x+40)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 4x+40)!=0

x∈R


We get rid of parentheses

2/5x-1/4x-40+10=0

We calculate fractions


8x/20x^2+(-5x)/20x^2-40+10=0

We add all the numbers together, and all the variables

8x/20x^2+(-5x)/20x^2-30=0

We multiply all the terms by the denominator


8x+(-5x)-30*20x^2=0

Wy multiply elements

-600x^2+8x+(-5x)=0

We get rid of parentheses

-600x^2+8x-5x=0

We add all the numbers together, and all the variables

-600x^2+3x=0

a = -600; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-600)·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*-600}=\frac{-6}{-1200}
=1/200
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-600}=\frac{0}{-1200}
=0
$