4^2+7^2=(2x)^2

Solution for 4^2+7^2=(2x)^2 equation:

4^2+7^2=(2x)^2

We move all terms to the left:

4^2+7^2-((2x)^2)=0

determiningTheFunctionDomain
-2x^2+4^2+7^2=0

We add all the numbers together, and all the variables

-2x^2+65=0

a = -2; b = 0; c = +65;
Δ = b2-4ac
Δ = 02-4·(-2)·65
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{130}}{2*-2}=\frac{0-2\sqrt{130}}{-4}
=-\frac{2\sqrt{130}}{-4}
=-\frac{\sqrt{130}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{130}}{2*-2}=\frac{0+2\sqrt{130}}{-4}
=\frac{2\sqrt{130}}{-4}
=\frac{\sqrt{130}}{-2}
$