What are the solutions to the equation t^2-4l=0 ?

The solutions to the equation t^2-4l=0 are t=2sqrt(l),t=-2sqrt(l)

Find the zeros of t^2-4l=0

The zeros of t^2-4l=0 are t=2sqrt(l),t=-2sqrt(l)

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t^{2} - 4 l = 0

t = 2 l, t = - 2 l

Solution steps

t^{2} - 4 l = 0

Move

4 l

to the right side

t^{2} = 4 l

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

t = 4 l, t = - 4 l

Simplify

4 l : 2 l

Simplify

- 4 l : - 2 l

t = 2 l, t = - 2 l

y = (y - 2)^{2}

4 x^{2} + 8 x = 6

x (2 x - 2) = 2

x^{2} + 14 x = - 13

3 x^{2} - 3 x + \frac{3}{4} = 0

What are the solutions to the equation t^2-4l=0 ?
The solutions to the equation t^2-4l=0 are t=2sqrt(l),t=-2sqrt(l)
Find the zeros of t^2-4l=0
The zeros of t^2-4l=0 are t=2sqrt(l),t=-2sqrt(l)