9/4x-3=x+4+5/4x

Solution for 9/4x-3=x+4+5/4x equation:

9/4x-3=x+4+5/4x

We move all terms to the left:

9/4x-3-(x+4+5/4x)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 4x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

9/4x-(x+5/4x+4)-3=0

We get rid of parentheses

9/4x-x-5/4x-4-3=0

We multiply all the terms by the denominator


-x*4x-4*4x-3*4x+9-5=0

We add all the numbers together, and all the variables

-x*4x-4*4x-3*4x+4=0

Wy multiply elements

-4x^2-16x-12x+4=0

We add all the numbers together, and all the variables

-4x^2-28x+4=0

a = -4; b = -28; c = +4;
Δ = b2-4ac
Δ = -282-4·(-4)·4
Δ = 848
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:

$\sqrt{\Delta}=\sqrt{848}=\sqrt{16*53}=\sqrt{16}*\sqrt{53}=4\sqrt{53}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{53}}{2*-4}=\frac{28-4\sqrt{53}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{53}}{2*-4}=\frac{28+4\sqrt{53}}{-8}
$