What are the solutions to the equation y^2=25+4+R ?

The solutions to the equation y^2=25+4+R are y=sqrt(R+29),y=-sqrt(R+29)

Find the zeros of y^2=25+4+R

The zeros of y^2=25+4+R are y=sqrt(R+29),y=-sqrt(R+29)

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y^{2} = 25 + 4 + R

y = R + 29, y = - R + 29

Solution steps

y^{2} = 25 + 4 + R

Add the numbers:

25 + 4 = 29

y^{2} = R + 29

For

x^{2} = f (a)

the solutions are

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

y = R + 29, y = - R + 29

\frac{1}{2} y - 7 = \frac{1}{4} y

3 x^{2} - 12 x = 69

26 = 13 x - 7 x^{2}

5 t^{2} - 50 t + 30 = 0

- 3 \leq 4 x < 12

What are the solutions to the equation y^2=25+4+R ?
The solutions to the equation y^2=25+4+R are y=sqrt(R+29),y=-sqrt(R+29)
Find the zeros of y^2=25+4+R
The zeros of y^2=25+4+R are y=sqrt(R+29),y=-sqrt(R+29)