(4+x)(4+x)=121

Solution for (4+x)(4+x)=121 equation:

(4+x)(4+x)=121

We move all terms to the left:

(4+x)(4+x)-(121)=0

We add all the numbers together, and all the variables

(x+4)(x+4)-121=0

We multiply parentheses ..

(+x^2+4x+4x+16)-121=0

We get rid of parentheses

x^2+4x+4x+16-121=0

We add all the numbers together, and all the variables

x^2+8x-105=0

a = 1; b = 8; c = -105;
Δ = b2-4ac
Δ = 82-4·1·(-105)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-22}{2*1}=\frac{-30}{2}
=-15
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+22}{2*1}=\frac{14}{2}
=7
$