(x+4)(x+4)=42

Solution for (x+4)(x+4)=42 equation:

(x+4)(x+4)=42

We move all terms to the left:

(x+4)(x+4)-(42)=0

We multiply parentheses ..

(+x^2+4x+4x+16)-42=0

We get rid of parentheses

x^2+4x+4x+16-42=0

We add all the numbers together, and all the variables

x^2+8x-26=0

a = 1; b = 8; c = -26;
Δ = b2-4ac
Δ = 82-4·1·(-26)
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{42}}{2*1}=\frac{-8-2\sqrt{42}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{42}}{2*1}=\frac{-8+2\sqrt{42}}{2}
$