2/5x+21=9/8x

Solution for 2/5x+21=9/8x equation:

2/5x+21=9/8x

We move all terms to the left:

2/5x+21-(9/8x)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 8x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

2/5x-(+9/8x)+21=0

We get rid of parentheses

2/5x-9/8x+21=0

We calculate fractions


16x/40x^2+(-45x)/40x^2+21=0

We multiply all the terms by the denominator


16x+(-45x)+21*40x^2=0

Wy multiply elements

840x^2+16x+(-45x)=0

We get rid of parentheses

840x^2+16x-45x=0

We add all the numbers together, and all the variables

840x^2-29x=0

a = 840; b = -29; c = 0;
Δ = b2-4ac
Δ = -292-4·840·0
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-29}{2*840}=\frac{0}{1680}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+29}{2*840}=\frac{58}{1680}
=29/840
$