(3/2)+(1/7)x=29

Solution for (3/2)+(1/7)x=29 equation:

(3/2)+(1/7)x=29

We move all terms to the left:

(3/2)+(1/7)x-(29)=0


Domain of the equation: 7)x!=0

x!=0/1

x!=0

x∈R


determiningTheFunctionDomain
(1/7)x-29+(3/2)=0

We add all the numbers together, and all the variables

(+1/7)x-29+(+3/2)=0

We multiply parentheses

x^2-29+(+3/2)=0

We get rid of parentheses

x^2-29+3/2=0

We multiply all the terms by the denominator


x^2*2+3-29*2=0

We add all the numbers together, and all the variables

x^2*2-55=0

Wy multiply elements

2x^2-55=0

a = 2; b = 0; c = -55;
Δ = b2-4ac
Δ = 02-4·2·(-55)
Δ = 440
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:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{110}}{2*2}=\frac{0-2\sqrt{110}}{4}
=-\frac{2\sqrt{110}}{4}
=-\frac{\sqrt{110}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{110}}{2*2}=\frac{0+2\sqrt{110}}{4}
=\frac{2\sqrt{110}}{4}
=\frac{\sqrt{110}}{2}
$