(X+4)3x=1+x

Solution for (X+4)3x=1+x equation:

(X+4)3X=1+X

We move all terms to the left:

(X+4)3X-(1+X)=0

We add all the numbers together, and all the variables

(X+4)3X-(X+1)=0

We multiply parentheses

3X^2+12X-(X+1)=0

We get rid of parentheses

3X^2+12X-X-1=0

We add all the numbers together, and all the variables

3X^2+11X-1=0

a = 3; b = 11; c = -1;
Δ = b2-4ac
Δ = 112-4·3·(-1)
Δ = 133
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{-(11)-\sqrt{133}}{2*3}=\frac{-11-\sqrt{133}}{6}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{133}}{2*3}=\frac{-11+\sqrt{133}}{6}
$