1/4x+41/8+1/2x=5/8

Solution for 1/4x+41/8+1/2x=5/8 equation:

1/4x+41/8+1/2x=5/8

We move all terms to the left:

1/4x+41/8+1/2x-(5/8)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

1/4x+1/2x+41/8-(+5/8)=0

We get rid of parentheses

1/4x+1/2x+41/8-5/8=0

We calculate fractions


(-80x^2+41)/512x^2+128x/512x^2+256x/512x^2=0

We multiply all the terms by the denominator


(-80x^2+41)+128x+256x=0

We add all the numbers together, and all the variables

(-80x^2+41)+384x=0

We get rid of parentheses

-80x^2+384x+41=0

a = -80; b = 384; c = +41;
Δ = b2-4ac
Δ = 3842-4·(-80)·41
Δ = 160576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160576}=\sqrt{64*2509}=\sqrt{64}*\sqrt{2509}=8\sqrt{2509}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(384)-8\sqrt{2509}}{2*-80}=\frac{-384-8\sqrt{2509}}{-160}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(384)+8\sqrt{2509}}{2*-80}=\frac{-384+8\sqrt{2509}}{-160}
$