(4x)^2+(3x)^2=(23)^2

Solution for (4x)^2+(3x)^2=(23)^2 equation:

(4x)^2+(3x)^2=(23)^2

We move all terms to the left:

(4x)^2+(3x)^2-((23)^2)=0

determiningTheFunctionDomain
4x^2+3x^2-23^2=0

We add all the numbers together, and all the variables

7x^2-529=0

a = 7; b = 0; c = -529;
Δ = b2-4ac
Δ = 02-4·7·(-529)
Δ = 14812
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:

$\sqrt{\Delta}=\sqrt{14812}=\sqrt{2116*7}=\sqrt{2116}*\sqrt{7}=46\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-46\sqrt{7}}{2*7}=\frac{0-46\sqrt{7}}{14}
=-\frac{46\sqrt{7}}{14}
=-\frac{23\sqrt{7}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+46\sqrt{7}}{2*7}=\frac{0+46\sqrt{7}}{14}
=\frac{46\sqrt{7}}{14}
=\frac{23\sqrt{7}}{7}
$