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1-4y^2=0

1-4y^2=0 equation

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Numerical solution:

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The solution

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       2    
1 - 4*y  = 0
$$1 - 4 y^{2} = 0$$
Detail solution
This equation is of the form
a*y^2 + b*y + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$y_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$y_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -4$$
$$b = 0$$
$$c = 1$$
, then
D = b^2 - 4 * a * c = 

(0)^2 - 4 * (-4) * (1) = 16

Because D > 0, then the equation has two roots.
y1 = (-b + sqrt(D)) / (2*a)

y2 = (-b - sqrt(D)) / (2*a)

or
$$y_{1} = - \frac{1}{2}$$
$$y_{2} = \frac{1}{2}$$
Vieta's Theorem
rewrite the equation
$$1 - 4 y^{2} = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$y^{2} - \frac{1}{4} = 0$$
$$p y + q + y^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = - \frac{1}{4}$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = 0$$
$$y_{1} y_{2} = - \frac{1}{4}$$
The graph
Rapid solution [src]
y1 = -1/2
$$y_{1} = - \frac{1}{2}$$
y2 = 1/2
$$y_{2} = \frac{1}{2}$$
y2 = 1/2
Sum and product of roots [src]
sum
-1/2 + 1/2
$$- \frac{1}{2} + \frac{1}{2}$$
=
0
$$0$$
product
-1 
---
2*2
$$- \frac{1}{4}$$
=
-1/4
$$- \frac{1}{4}$$
-1/4
Numerical answer [src]
y1 = 0.5
y2 = -0.5
y2 = -0.5
The graph
1-4y^2=0 equation