What are the solutions to the equation 5(x-3)^2+4=129 ?

The solutions to the equation 5(x-3)^2+4=129 are x=8,x=-2

Find the zeros of 5(x-3)^2+4=129

The zeros of 5(x-3)^2+4=129 are x=8,x=-2

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5 (x - 3)^{2} + 4 = 129

x = 8, x = - 2

Solution steps

5 (x - 3)^{2} + 4 = 129

Move

4

to the right side

5 (x - 3)^{2} = 125

Divide both sides by

5

(x - 3)^{2} = 25

For

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

the solutions are

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

Solve

x - 3 = 25 : x = 8

Solve

x - 3 = - 25 : x = - 2

The solutions to the quadratic equation are:

x = 8, x = - 2

3 x^{2} + 12 x + 10 = 0

\frac{2 x - 5}{9} = 3

∣ 5 x + 2 ∣ \leq 3

2 (4 k^{2} - 2 k) - 3 (- 6 k^{2} + 4) + 2 k (k - 1)

45 a^{4} - 30 a^{3} + 5 a^{2}

What are the solutions to the equation 5(x-3)^2+4=129 ?
The solutions to the equation 5(x-3)^2+4=129 are x=8,x=-2
Find the zeros of 5(x-3)^2+4=129
The zeros of 5(x-3)^2+4=129 are x=8,x=-2