x^2+(4x)^2=10.5^2

Solution for x^2+(4x)^2=10.5^2 equation:

x^2+(4x)^2=10.5^2

We move all terms to the left:

x^2+(4x)^2-(10.5^2)=0

We add all the numbers together, and all the variables

5x^2-110.25=0

a = 5; b = 0; c = -110.25;
Δ = b2-4ac
Δ = 02-4·5·(-110.25)
Δ = 2205
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:

$\sqrt{\Delta}=\sqrt{2205}=\sqrt{441*5}=\sqrt{441}*\sqrt{5}=21\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-21\sqrt{5}}{2*5}=\frac{0-21\sqrt{5}}{10}
=-\frac{21\sqrt{5}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+21\sqrt{5}}{2*5}=\frac{0+21\sqrt{5}}{10}
=\frac{21\sqrt{5}}{10}
$