What are the solutions to the equation (w+1)^2-48=0 ?

The solutions to the equation (w+1)^2-48=0 are w=4sqrt(3)-1,w=-4sqrt(3)-1

Find the zeros of (w+1)^2-48=0

The zeros of (w+1)^2-48=0 are w=4sqrt(3)-1,w=-4sqrt(3)-1

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(w + 1)^{2} - 48 = 0

w = 43 - 1, w = - 43 - 1

Solution steps

(w + 1)^{2} - 48 = 0

Move

48

to the right side

(w + 1)^{2} = 48

For

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

the solutions are

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

Solve

w + 1 = 48 : w = 43 - 1

Solve

w + 1 = - 48 : w = - 43 - 1

The solutions to the quadratic equation are:

w = 43 - 1, w = - 43 - 1

(8)^{2}

lo g_{25} (5) = w

x^{2} + 8 x = 1

7 x^{2} - 10 x + 9 = 5 - 2 x

\frac{a ^{2} - 4 a c + 3 b c}{a ^{2} - ab + b c - a c} + \frac{a + 3 b}{b - a} + \frac{a + 2 c}{a - c}

What are the solutions to the equation (w+1)^2-48=0 ?
The solutions to the equation (w+1)^2-48=0 are w=4sqrt(3)-1,w=-4sqrt(3)-1
Find the zeros of (w+1)^2-48=0
The zeros of (w+1)^2-48=0 are w=4sqrt(3)-1,w=-4sqrt(3)-1