((4x)^2)+((3x)^2)=20.25

Solution for ((4x)^2)+((3x)^2)=20.25 equation:

((4x)^2)+((3x)^2)=20.25

We move all terms to the left:

((4x)^2)+((3x)^2)-(20.25)=0

determiningTheFunctionDomain
4x^2+3x^2-(20.25)=0

We add all the numbers together, and all the variables

7x^2-20.25=0

a = 7; b = 0; c = -20.25;
Δ = b2-4ac
Δ = 02-4·7·(-20.25)
Δ = 567
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:

$\sqrt{\Delta}=\sqrt{567}=\sqrt{81*7}=\sqrt{81}*\sqrt{7}=9\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-9\sqrt{7}}{2*7}=\frac{0-9\sqrt{7}}{14}
=-\frac{9\sqrt{7}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+9\sqrt{7}}{2*7}=\frac{0+9\sqrt{7}}{14}
=\frac{9\sqrt{7}}{14}
$