(2x+3)2+(3x)2=20

Solution for (2x+3)2+(3x)2=20 equation:

(2x+3)2+(3x)2=20

We move all terms to the left:

(2x+3)2+(3x)2-(20)=0

We add all the numbers together, and all the variables

3x^2+(2x+3)2-20=0

We multiply parentheses

3x^2+4x+6-20=0

We add all the numbers together, and all the variables

3x^2+4x-14=0

a = 3; b = 4; c = -14;
Δ = b2-4ac
Δ = 42-4·3·(-14)
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{46}}{2*3}=\frac{-4-2\sqrt{46}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{46}}{2*3}=\frac{-4+2\sqrt{46}}{6}
$