How can I solve $5^{2x} + 4(5^x) - 5 = 0$? This is a math problem I'm currently working on.
$$5^{2x} + 4(5^x) - 5 = 0$$
I've used logarithm to try solve the problem. Here's what I've done so far: 
\begin{align}5^{2x} + 4(5^x) - 5 &= 0\\
5^{2x} + 5^x &= \frac{5}{4}\\
\log_5{2x} + \log_5{x} &= \log_5\left(\frac{5}{4}\right)\\
\log_5{2x^2} &= \log_5\left(\frac{5}{4}\right)\\
5^{2x^2} &= 5^\frac{5}{4}\\
(2x^2)\log5 &= \frac{5}{4}\log5\\
2x^2 &= \frac{5}{4}\\
2x^2 &= 1.25\\
x^2 &= 0.625\\
\sqrt {x^2} &= \sqrt {0.625}\\
x &= \frac{\sqrt {10}}{4}\end{align}
Substituting the value for $x$ into the equation doesn't equate it to zero. I've tried several different ways but I have still not come up with a correct answer. 

$\require{enclose}5^{2x}+4(5)^{x}-5=0 \enclose{updiagonalstrike}{\implies} 5^{2x}+5^{x}=\frac{5}{4}$
Instead:
Let $y=5^x\quad$ noting that $y\geq 0$, then we have:
$y^2+4y-5=0\implies (y+5)(y-1)=0$
$\implies y=-5\quad\text{reject since}\quad y\geq 0\quad \text{or}\quad y=1\implies 5^x=1\implies x=0$
