A(x2+4x)=13

Solution for A(x2+4x)=13 equation:

(A2+4A)=13

We move all terms to the left:

(A2+4A)-(13)=0

We add all the numbers together, and all the variables

(+A^2+4A)-13=0

We get rid of parentheses

A^2+4A-13=0

a = 1; b = 4; c = -13;
Δ = b2-4ac
Δ = 42-4·1·(-13)
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{17}}{2*1}=\frac{-4-2\sqrt{17}}{2}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{17}}{2*1}=\frac{-4+2\sqrt{17}}{2}
$