What are the solutions to the equation 9-5(t+1)^2=4 ?

The solutions to the equation 9-5(t+1)^2=4 are t=0,t=-2

Find the zeros of 9-5(t+1)^2=4

The zeros of 9-5(t+1)^2=4 are t=0,t=-2

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9 - 5 (t + 1)^{2} = 4

t = 0, t = - 2

Solution steps

9 - 5 (t + 1)^{2} = 4

Move

9

to the right side

- 5 (t + 1)^{2} = - 5

Divide both sides by

- 5

(t + 1)^{2} = 1

For

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

the solutions are

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

Solve

t + 1 = 1 : t = 0

Solve

t + 1 = - 1 : t = - 2

The solutions to the quadratic equation are:

t = 0, t = - 2

18 u - 5 u + 29 = 6 u - 27

\frac{b ^{3}}{b ^{7}}

m^{2} + 2 m - 3 = 0

\frac{π}{8} + \frac{π}{4}

4^{\frac{1}{6}}

What are the solutions to the equation 9-5(t+1)^2=4 ?
The solutions to the equation 9-5(t+1)^2=4 are t=0,t=-2
Find the zeros of 9-5(t+1)^2=4
The zeros of 9-5(t+1)^2=4 are t=0,t=-2