What are the solutions to the equation (w+1/2)^2+4=21 ?

The solutions to the equation (w+1/2)^2+4=21 are w=sqrt(17)-1/2 ,w=-sqrt(17)-1/2

Find the zeros of (w+1/2)^2+4=21

The zeros of (w+1/2)^2+4=21 are w=sqrt(17)-1/2 ,w=-sqrt(17)-1/2

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(w + \frac{1}{2})^{2} + 4 = 21

w = 17 - \frac{1}{2}, w = - 17 - \frac{1}{2}

Decimal

w = 3.62310 \dots, w = - 4.62310 \dots

Solution steps

(w + \frac{1}{2})^{2} + 4 = 21

Move

4

to the right side

(w + \frac{1}{2})^{2} = 17

For

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

the solutions are

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

Solve

w + \frac{1}{2} = 17 : w = 17 - \frac{1}{2}

Solve

w + \frac{1}{2} = - 17 : w = - 17 - \frac{1}{2}

The solutions to the quadratic equation are:

w = 17 - \frac{1}{2}, w = - 17 - \frac{1}{2}

3 t^{2} - 6 t

3 (x - 4) (x - 1) = 0

y^{2} - 4 = 12

w^{2} + 12 = 0

(y + 1)^{2} = 9

What are the solutions to the equation (w+1/2)^2+4=21 ?
The solutions to the equation (w+1/2)^2+4=21 are w=sqrt(17)-1/2 ,w=-sqrt(17)-1/2
Find the zeros of (w+1/2)^2+4=21
The zeros of (w+1/2)^2+4=21 are w=sqrt(17)-1/2 ,w=-sqrt(17)-1/2