(2x)^2+(3x)^2=39x

Solution for (2x)^2+(3x)^2=39x equation:

(2x)^2+(3x)^2=39x

We move all terms to the left:

(2x)^2+(3x)^2-(39x)=0

We add all the numbers together, and all the variables

5x^2-39x=0

a = 5; b = -39; c = 0;
Δ = b2-4ac
Δ = -392-4·5·0
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-39}{2*5}=\frac{0}{10}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+39}{2*5}=\frac{78}{10}
=7+4/5
$