(3/2)+(1/7)x=2

Solution for (3/2)+(1/7)x=2 equation:

(3/2)+(1/7)x=2

We move all terms to the left:

(3/2)+(1/7)x-(2)=0


Domain of the equation: 7)x!=0

x!=0/1

x!=0

x∈R


determiningTheFunctionDomain
(1/7)x-2+(3/2)=0

We add all the numbers together, and all the variables

(+1/7)x-2+(+3/2)=0

We multiply parentheses

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

We get rid of parentheses

x^2-2+3/2=0

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

x^2*2-1=0

Wy multiply elements

2x^2-1=0

a = 2; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·2·(-1)
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*2}=\frac{0-2\sqrt{2}}{4}
=-\frac{2\sqrt{2}}{4}
=-\frac{\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*2}=\frac{0+2\sqrt{2}}{4}
=\frac{2\sqrt{2}}{4}
=\frac{\sqrt{2}}{2}
$