What are the solutions to the equation (y+1)^2= 3/2 ?

The solutions to the equation (y+1)^2= 3/2 are y=sqrt(3/2)-1,y=-sqrt(3/2)-1

Find the zeros of (y+1)^2= 3/2

The zeros of (y+1)^2= 3/2 are y=sqrt(3/2)-1,y=-sqrt(3/2)-1

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(y + 1)^{2} = \frac{3}{2}

y = \frac{3}{2} - 1, y = - \frac{3}{2} - 1

Decimal Notation

y = 0.22474 \dots, y = - 2.22474 \dots

Solution steps

(y + 1)^{2} = \frac{3}{2}

For

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

the solutions are

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

Solve

y + 1 = \frac{3}{2} : y = \frac{3}{2} - 1

Solve

y + 1 = - \frac{3}{2} : y = - \frac{3}{2} - 1

The solutions to the quadratic equation are:

y = \frac{3}{2} - 1, y = - \frac{3}{2} - 1

(2 x - 5)^{2} \geq 72 + (- 4 + x)^{2}

\frac{16}{9}

\frac{c ^{2}}{20} - \frac{c}{4} + \frac{1}{5} = 0

t^{2} - t + 2 = 0

3 x^{2} - 5 x - 8 = 0

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