What are the solutions to the equation ((3y)/4)^2+y^2=225 ?

The solutions to the equation ((3y)/4)^2+y^2=225 are y=12,y=-12

Find the zeros of ((3y)/4)^2+y^2=225

The zeros of ((3y)/4)^2+y^2=225 are y=12,y=-12

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(\frac{3 y}{4})^{2} + y^{2} = 225

y = 12, y = - 12

Solution steps

(\frac{3 y}{4})^{2} + y^{2} = 225

Expand

(\frac{3 y}{4})^{2} + y^{2} : \frac{25 y ^{2}}{16}

\frac{25 y ^{2}}{16} = 225

Move

225

to the left side

\frac{25 y ^{2}}{16} - 225 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot \frac{25}{16} ( - 225 )}{2 \cdot \frac{25}{16}}

0^{2} - 4 \cdot \frac{25}{16} (- 225) = \frac{75}{2}

y_{1, 2} = \frac{- 0 \pm \frac{75}{2}}{2 \cdot \frac{25}{16}}

Separate the solutions

y_{1} = \frac{- 0 + \frac{75}{2}}{2 \cdot \frac{25}{16}}, y_{2} = \frac{- 0 - \frac{75}{2}}{2 \cdot \frac{25}{16}}

y = \frac{- 0 + \frac{75}{2}}{2 \frac{25}{16}} : 12

y = \frac{- 0 - \frac{75}{2}}{2 \frac{25}{16}} : - 12

The solutions to the quadratic equation are:

y = 12, y = - 12

3 y^{2} + 14 y = 5

S = U - N (N)

(x + 7)^{2} = - 4

x^{2} - 1.4 x + 0.4 = 0

x^{2} + 9 x + 20 = 0

What are the solutions to the equation ((3y)/4)^2+y^2=225 ?
The solutions to the equation ((3y)/4)^2+y^2=225 are y=12,y=-12
Find the zeros of ((3y)/4)^2+y^2=225
The zeros of ((3y)/4)^2+y^2=225 are y=12,y=-12