(3/2)x+(5/4)=3

Solution for (3/2)x+(5/4)=3 equation:

(3/2)x+(5/4)=3

We move all terms to the left:

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


Domain of the equation: 2)x!=0

x!=0/1

x!=0

x∈R


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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

3x^2*4-7=0

Wy multiply elements

12x^2-7=0

a = 12; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·12·(-7)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{21}}{2*12}=\frac{0-4\sqrt{21}}{24}
=-\frac{4\sqrt{21}}{24}
=-\frac{\sqrt{21}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{21}}{2*12}=\frac{0+4\sqrt{21}}{24}
=\frac{4\sqrt{21}}{24}
=\frac{\sqrt{21}}{6}
$