What are the solutions to the equation 4=3+(y-5)^2 ?

The solutions to the equation 4=3+(y-5)^2 are y=6,y=4

Find the zeros of 4=3+(y-5)^2

The zeros of 4=3+(y-5)^2 are y=6,y=4

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4 = 3 + (y - 5)^{2}

y = 6, y = 4

Solution steps

4 = 3 + (y - 5)^{2}

Switch sides

3 + (y - 5)^{2} = 4

Move

3

to the right side

(y - 5)^{2} = 1

For

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

the solutions are

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

Solve

y - 5 = 1 : y = 6

Solve

y - 5 = - 1 : y = 4

The solutions to the quadratic equation are:

y = 6, y = 4

0 = - 16 t^{2} + 16 t + 320

8 - p^{2} = p^{2}

x - 1 = x^{2} - 1

x^{2} + 160 = 0

s^{2} 5 + s + 5 = 0

What are the solutions to the equation 4=3+(y-5)^2 ?
The solutions to the equation 4=3+(y-5)^2 are y=6,y=4
Find the zeros of 4=3+(y-5)^2
The zeros of 4=3+(y-5)^2 are y=6,y=4