x^2+4x^2=41

Solution for x^2+4x^2=41 equation:

x^2+4x^2=41

We move all terms to the left:

x^2+4x^2-(41)=0

We add all the numbers together, and all the variables

5x^2-41=0

a = 5; b = 0; c = -41;
Δ = b2-4ac
Δ = 02-4·5·(-41)
Δ = 820
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:

$\sqrt{\Delta}=\sqrt{820}=\sqrt{4*205}=\sqrt{4}*\sqrt{205}=2\sqrt{205}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{205}}{2*5}=\frac{0-2\sqrt{205}}{10}
=-\frac{2\sqrt{205}}{10}
=-\frac{\sqrt{205}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{205}}{2*5}=\frac{0+2\sqrt{205}}{10}
=\frac{2\sqrt{205}}{10}
=\frac{\sqrt{205}}{5}
$