(3/4)x+(5/4)=4x

Solution for (3/4)x+(5/4)=4x equation:

(3/4)x+(5/4)=4x

We move all terms to the left:

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


Domain of the equation: 4)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3x^2-4x+(+5/4)=0

We get rid of parentheses

3x^2-4x+5/4=0

We multiply all the terms by the denominator


3x^2*4-4x*4+5=0

Wy multiply elements

12x^2-16x+5=0

a = 12; b = -16; c = +5;
Δ = b2-4ac
Δ = -162-4·12·5
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*12}=\frac{12}{24}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*12}=\frac{20}{24}
=5/6
$