(2x+3x)2-(x-4)(x+4)=21

Solution for (2x+3x)2-(x-4)(x+4)=21 equation:

(2x+3x)2-(x-4)(x+4)=21

We move all terms to the left:

(2x+3x)2-(x-4)(x+4)-(21)=0

We add all the numbers together, and all the variables

(+5x)2-(x-4)(x+4)-21=0

We use the square of the difference formula

x^2+(+5x)2+16-21=0

We multiply parentheses

x^2+10x+16-21=0

We add all the numbers together, and all the variables

x^2+10x-5=0

a = 1; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·1·(-5)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{30}}{2*1}=\frac{-10-2\sqrt{30}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{30}}{2*1}=\frac{-10+2\sqrt{30}}{2}
$