0.5x+(4/3x)=x+5

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Solution for 0.5x+(4/3x)=x+5 equation:



0.5x+(4/3x)=x+5
We move all terms to the left:
0.5x+(4/3x)-(x+5)=0
Domain of the equation: 3x)!=0
x!=0/1
x!=0
x∈R
We add all the numbers together, and all the variables
0.5x+(+4/3x)-(x+5)=0
We get rid of parentheses
0.5x+4/3x-x-5=0
We multiply all the terms by the denominator
(0.5x)*3x-x*3x-5*3x+4=0
We add all the numbers together, and all the variables
(+0.5x)*3x-x*3x-5*3x+4=0
We multiply parentheses
0x^2-x*3x-5*3x+4=0
Wy multiply elements
0x^2-3x^2-15x+4=0
We add all the numbers together, and all the variables
-2x^2-15x+4=0
a = -2; b = -15; c = +4;
Δ = b2-4ac
Δ = -152-4·(-2)·4
Δ = 257
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{257}}{2*-2}=\frac{15-\sqrt{257}}{-4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{257}}{2*-2}=\frac{15+\sqrt{257}}{-4} $

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