What are the solutions to the equation (4t+1)^2=49 ?

The solutions to the equation (4t+1)^2=49 are t= 3/2 ,t=-2

Find the zeros of (4t+1)^2=49

The zeros of (4t+1)^2=49 are t= 3/2 ,t=-2

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(4 t + 1)^{2} = 49

t = \frac{3}{2}, t = - 2

Decimal Notation

t = 1.5, t = - 2

Solution steps

(4 t + 1)^{2} = 49

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

4 t + 1 = 49 : t = \frac{3}{2}

Solve

4 t + 1 = - 49 : t = - 2

The solutions to the quadratic equation are:

t = \frac{3}{2}, t = - 2

4 (2 - x) < - 2 and \frac{x - 3}{5} \leq - 1

- 264 = - 8 (5 k - 2)

(\frac{9}{2})^{- 1}

c^{4} - 27 c

2 x^{2} - 2 x - 6 = 0

What are the solutions to the equation (4t+1)^2=49 ?
The solutions to the equation (4t+1)^2=49 are t= 3/2 ,t=-2
Find the zeros of (4t+1)^2=49
The zeros of (4t+1)^2=49 are t= 3/2 ,t=-2