What are the solutions to the equation ((-5-3)^2)/9-((y-4)^2)/4 =1 ?

The solutions to the equation ((-5-3)^2)/9-((y-4)^2)/4 =1 are y=(2sqrt(55))/3+4,y=-(2sqrt(55))/3+4

Find the zeros of ((-5-3)^2)/9-((y-4)^2)/4 =1

The zeros of ((-5-3)^2)/9-((y-4)^2)/4 =1 are y=(2sqrt(55))/3+4,y=-(2sqrt(55))/3+4

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\frac{( - 5 - 3 ) ^{2}}{9} - \frac{( y - 4 ) ^{2}}{4} = 1

y = \frac{2 55}{3} + 4, y = - \frac{2 55}{3} + 4

Decimal

y = 8.94413 \dots, y = - 0.94413 \dots

Solution steps

\frac{( - 5 - 3 ) ^{2}}{9} - \frac{( y - 4 ) ^{2}}{4} = 1

Move

\frac{( - 5 - 3 ) ^{2}}{9}

to the right side

- \frac{( y - 4 ) ^{2}}{4} = 1 - \frac{64}{9}

Simplify

1 - \frac{64}{9} : - \frac{55}{9}

- \frac{( y - 4 ) ^{2}}{4} = - \frac{55}{9}

Multiply both sides by

4

- (y - 4)^{2} = - \frac{220}{9}

Divide both sides by

- 1

(y - 4)^{2} = \frac{220}{9}

For

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

the solutions are

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

Solve

y - 4 = \frac{220}{9} : y = \frac{2 55}{3} + 4

Solve

y - 4 = - \frac{220}{9} : y = - \frac{2 55}{3} + 4

The solutions to the quadratic equation are:

y = \frac{2 55}{3} + 4, y = - \frac{2 55}{3} + 4

c, e = m \cdot c^{2}

28 = y^{2}

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

2 x^{2} = 10 x - 11

5 x^{2} = 25 x

What are the solutions to the equation ((-5-3)^2)/9-((y-4)^2)/4 =1 ?
The solutions to the equation ((-5-3)^2)/9-((y-4)^2)/4 =1 are y=(2sqrt(55))/3+4,y=-(2sqrt(55))/3+4
Find the zeros of ((-5-3)^2)/9-((y-4)^2)/4 =1
The zeros of ((-5-3)^2)/9-((y-4)^2)/4 =1 are y=(2sqrt(55))/3+4,y=-(2sqrt(55))/3+4