(x+3)(x+3)=40

Solution for (x+3)(x+3)=40 equation:

(x+3)(x+3)=40

We move all terms to the left:

(x+3)(x+3)-(40)=0

We multiply parentheses ..

(+x^2+3x+3x+9)-40=0

We get rid of parentheses

x^2+3x+3x+9-40=0

We add all the numbers together, and all the variables

x^2+6x-31=0

a = 1; b = 6; c = -31;
Δ = b2-4ac
Δ = 62-4·1·(-31)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{10}}{2*1}=\frac{-6-4\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{10}}{2*1}=\frac{-6+4\sqrt{10}}{2}
$