(1/3)(x+4)=2x+5

Solution for (1/3)(x+4)=2x+5 equation:

(1/3)(x+4)=2x+5

We move all terms to the left:

(1/3)(x+4)-(2x+5)=0


Domain of the equation: 3)(x+4)!=0

x∈R


We add all the numbers together, and all the variables

(+1/3)(x+4)-(2x+5)=0

We get rid of parentheses

(+1/3)(x+4)-2x-5=0

We multiply parentheses ..

(+x^2+1/3*4)-2x-5=0

We multiply all the terms by the denominator


(+x^2+1-2x*3*4)-5*3*4)=0

We add all the numbers together, and all the variables

(+x^2+1-2x*3*4)=0

We get rid of parentheses

x^2-2x*3*4+1=0

Wy multiply elements

x^2-24x*4+1=0

Wy multiply elements

x^2-96x+1=0

a = 1; b = -96; c = +1;
Δ = b2-4ac
Δ = -962-4·1·1
Δ = 9212
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:

$\sqrt{\Delta}=\sqrt{9212}=\sqrt{196*47}=\sqrt{196}*\sqrt{47}=14\sqrt{47}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-14\sqrt{47}}{2*1}=\frac{96-14\sqrt{47}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+14\sqrt{47}}{2*1}=\frac{96+14\sqrt{47}}{2}
$