(2/3)x+(3/4)=51

Solution for (2/3)x+(3/4)=51 equation:

(2/3)x+(3/4)=51

We move all terms to the left:

(2/3)x+(3/4)-(51)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R


determiningTheFunctionDomain
(2/3)x-51+(3/4)=0

We add all the numbers together, and all the variables

(+2/3)x-51+(+3/4)=0

We multiply parentheses

2x^2-51+(+3/4)=0

We get rid of parentheses

2x^2-51+3/4=0

We multiply all the terms by the denominator


2x^2*4+3-51*4=0

We add all the numbers together, and all the variables

2x^2*4-201=0

Wy multiply elements

8x^2-201=0

a = 8; b = 0; c = -201;
Δ = b2-4ac
Δ = 02-4·8·(-201)
Δ = 6432
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:

$\sqrt{\Delta}=\sqrt{6432}=\sqrt{16*402}=\sqrt{16}*\sqrt{402}=4\sqrt{402}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{402}}{2*8}=\frac{0-4\sqrt{402}}{16}
=-\frac{4\sqrt{402}}{16}
=-\frac{\sqrt{402}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{402}}{2*8}=\frac{0+4\sqrt{402}}{16}
=\frac{4\sqrt{402}}{16}
=\frac{\sqrt{402}}{4}
$