What are the solutions to the equation 64+(25/3)^2=y^2 ?

The solutions to the equation 64+(25/3)^2=y^2 are y=(sqrt(1201))/3 ,y=-(sqrt(1201))/3

Find the zeros of 64+(25/3)^2=y^2

The zeros of 64+(25/3)^2=y^2 are y=(sqrt(1201))/3 ,y=-(sqrt(1201))/3

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64 + (\frac{25}{3})^{2} = y^{2}

y = \frac{1201}{3}, y = - \frac{1201}{3}

Decimal Notation

y = 11.55181 \dots, y = - 11.55181 \dots

Solution steps

64 + (\frac{25}{3})^{2} = y^{2}

Switch sides

y^{2} = 64 + (\frac{25}{3})^{2}

Simplify

64 + (\frac{25}{3})^{2} : \frac{1201}{9}

For

x^{2} = f (a)

the solutions are

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

y = \frac{1201}{9}, y = - \frac{1201}{9}

\frac{1201}{9} = \frac{1201}{3}

- \frac{1201}{9} = - \frac{1201}{3}

y = \frac{1201}{3}, y = - \frac{1201}{3}

- x^{2} + 7 x - 6 = 0

4 (4 - 6 t) + (1 + 9 t) + 2 (2 + 3 t) = 1

4 (\frac{2}{3})

w^{2} + 14 w + 49

\frac{5 x - 2}{3} - \frac{x - 8}{4} > \frac{x + 14}{2} - 2

What are the solutions to the equation 64+(25/3)^2=y^2 ?
The solutions to the equation 64+(25/3)^2=y^2 are y=(sqrt(1201))/3 ,y=-(sqrt(1201))/3
Find the zeros of 64+(25/3)^2=y^2
The zeros of 64+(25/3)^2=y^2 are y=(sqrt(1201))/3 ,y=-(sqrt(1201))/3