25x^2+4(2x)^2-100=0

Solution for 25x^2+4(2x)^2-100=0 equation:

25x^2+4(2x)^2-100=0

We add all the numbers together, and all the variables

67x^2-100=0

a = 67; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·67·(-100)
Δ = 26800
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:

$\sqrt{\Delta}=\sqrt{26800}=\sqrt{400*67}=\sqrt{400}*\sqrt{67}=20\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{67}}{2*67}=\frac{0-20\sqrt{67}}{134}
=-\frac{20\sqrt{67}}{134}
=-\frac{10\sqrt{67}}{67}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{67}}{2*67}=\frac{0+20\sqrt{67}}{134}
=\frac{20\sqrt{67}}{134}
=\frac{10\sqrt{67}}{67}
$