(1/7)(x+4)=5

Solution for (1/7)(x+4)=5 equation:

(1/7)(x+4)=5

We move all terms to the left:

(1/7)(x+4)-(5)=0


Domain of the equation: 7)(x+4)!=0

x∈R


We add all the numbers together, and all the variables

(+1/7)(x+4)-5=0

We multiply parentheses ..

(+x^2+1/7*4)-5=0

We multiply all the terms by the denominator


(+x^2+1-5*7*4)=0

We get rid of parentheses

x^2+1-5*7*4=0

We add all the numbers together, and all the variables

x^2-139=0

a = 1; b = 0; c = -139;
Δ = b2-4ac
Δ = 02-4·1·(-139)
Δ = 556
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:

$\sqrt{\Delta}=\sqrt{556}=\sqrt{4*139}=\sqrt{4}*\sqrt{139}=2\sqrt{139}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{139}}{2*1}=\frac{0-2\sqrt{139}}{2}
=-\frac{2\sqrt{139}}{2}
=-\sqrt{139}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{139}}{2*1}=\frac{0+2\sqrt{139}}{2}
=\frac{2\sqrt{139}}{2}
=\sqrt{139}
$