4(z+3)=(1/3)(12z+36)

Solution for 4(z+3)=(1/3)(12z+36) equation:

4(z+3)=(1/3)(12z+36)

We move all terms to the left:

4(z+3)-((1/3)(12z+36))=0


Domain of the equation: 3)(12z+36))!=0

z∈R


We add all the numbers together, and all the variables

4(z+3)-((+1/3)(12z+36))=0

We multiply parentheses

4z-((+1/3)(12z+36))+12=0

We multiply parentheses ..

-((+12z^2+1/3*36))+4z+12=0

We multiply all the terms by the denominator


-((+12z^2+1+4z*3*36))+12*3*36))=0


We calculate terms in parentheses: -((+12z^2+1+4z*3*36)), so:

(+12z^2+1+4z*3*36)

We get rid of parentheses

12z^2+4z*3*36+1

Wy multiply elements

12z^2+432z*3+1

Wy multiply elements

12z^2+1296z+1

Back to the equation:

-(12z^2+1296z+1)


We add all the numbers together, and all the variables

-(12z^2+1296z+1)=0

We get rid of parentheses

-12z^2-1296z-1=0

a = -12; b = -1296; c = -1;
Δ = b2-4ac
Δ = -12962-4·(-12)·(-1)
Δ = 1679568
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1679568}=\sqrt{16*104973}=\sqrt{16}*\sqrt{104973}=4\sqrt{104973}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1296)-4\sqrt{104973}}{2*-12}=\frac{1296-4\sqrt{104973}}{-24}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1296)+4\sqrt{104973}}{2*-12}=\frac{1296+4\sqrt{104973}}{-24}
$