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

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## 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 = b2-4acΔ = 02-4·8·(-201)Δ = 6432The delta value is higher than zero, so the equation has two solutionsWe use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$The end solution:
$\sqrt{\Delta}=\sqrt{6432}=\sqrt{16*402}=\sqrt{16}*\sqrt{402}=4\sqrt{402}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{402}}{2*8}=\frac{0-4\sqrt{402}}{16} =-\frac{4\sqrt{402}}{16} =-\frac{\sqrt{402}}{4}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{402}}{2*8}=\frac{0+4\sqrt{402}}{16} =\frac{4\sqrt{402}}{16} =\frac{\sqrt{402}}{4}$

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