How to prove $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$ So, I was watching this video by blackpenredpen where he mentions that $$\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$$ so I wanted to try and prove it myself.  
Let $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=x$
But$\sqrt {a-\sqrt {a+\underbrace{\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}_x}}=x$
$\therefore \sqrt {a-\sqrt {a+x}}=x$
$a-\sqrt {a+x}=x^2$
$x^2-a=-\sqrt {a+x}$
$x^4-2ax^2+a^2=a+x$
$x^4-2ax^2-x+a^2-a=0$
Note that this is of the form $y^4+py^2+qy+r=0$ so we can use Ferrari-Cardano.   
We need to find a $z$ such that $(2z-p)y^2-qy+(z^2-r)$ has a discriminant of $0$. The discriminant of $(2z-p)y^2-qy+(z^2-r)$ is equal to $q^2 - 4(2z - p)(z^2 - r),$ which simplifies to $8z^3 - 4pz^2 - 8rz + (4pr - q^2) = 0$ 
Substituting values from $x^4-2ax^2-x+a^2-a=0$ into $8z^3 - 4pz^2 - 8rz + (4pr - q^2) = 0$ gives us
$8z^3-4\cdot(-2a)\cdot z^2-8\cdot(-a)\cdot z+\left(4\cdot (-2a) \cdot (-a) - (-1)^2 \right)=0 \implies 8z^3+8az^2+8az+(8a^2-1)=0$ 
Using Cardano's formula, or in my case Wolfram Alpha, we get that $$z_1 = \frac {\sqrt [3]{-16 a^3 - 144 a^2 + 3 \sqrt 3 \sqrt {256 a^5 + 512 a^4 + 224 a^3 - 288 a^2 + 27} + 27}}{6\sqrt[3]2} - \frac {192 a - 64 a^2}{48\cdot 2^{\frac 23} \sqrt [3]{-16 a^3 - 144 a^2 + 3\sqrt 3 \sqrt {256 a^5 + 512 a^4 + 224 a^3 - 288 a^2 + 27} + 27}} - \frac a3$$ 
I simply can not solve that quintic and continue as it is already too cluttered. Was there a mistake in my problem or is there any other way to do it? Also, I am sincerely sorry but I am not sure how to tag this question.

Edit $1:$ 
After Ross Millikan's answer, I snooped around in the comment section of the video and found someone who found that it is true using alternating root series. Was his proof correct as $\frac {\sqrt {4a-3}-1}2$ does not seem to have real values for $a \lt \frac 34$? Thank you!

It is CORRECT!
$x^4-2ax^2-x+a^2-a=0$ can be factorised as $(x^2-x-1)*(x^2+x+1-a)$. You'll get the desired answer from the right factor $(x^2+x+1-a)$.
