What are the solutions to the equation 25(3-5x)^2-4=25 ?

The solutions to the equation 25(3-5x)^2-4=25 are x=(-sqrt(29)+15)/(25),x=(sqrt(29)+15)/(25)

Find the zeros of 25(3-5x)^2-4=25

The zeros of 25(3-5x)^2-4=25 are x=(-sqrt(29)+15)/(25),x=(sqrt(29)+15)/(25)

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25 (3 - 5 x)^{2} - 4 = 25

x = \frac{- 29 + 15}{25}, x = \frac{29 + 15}{25}

Decimal Notation

x = 0.38459 \dots, x = 0.81540 \dots

Solution steps

25 (3 - 5 x)^{2} - 4 = 25

Move

4

to the right side

25 (3 - 5 x)^{2} = 29

Divide both sides by

25

(3 - 5 x)^{2} = \frac{29}{25}

For

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

the solutions are

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

Solve

3 - 5 x = \frac{29}{25} : x = \frac{- 29 + 15}{25}

Solve

3 - 5 x = - \frac{29}{25} : x = \frac{29 + 15}{25}

The solutions to the quadratic equation are:

x = \frac{- 29 + 15}{25}, x = \frac{29 + 15}{25}

(x - 9)^{2} - 1 = 0

x, f (a + 1) = 5 x^{2} - x + 2

2 x^{2} - 3 x + 2

4^{2} = 2^{2} + b^{2}

(5 x - 4)^{2} - 1 = 0

What are the solutions to the equation 25(3-5x)^2-4=25 ?
The solutions to the equation 25(3-5x)^2-4=25 are x=(-sqrt(29)+15)/(25),x=(sqrt(29)+15)/(25)
Find the zeros of 25(3-5x)^2-4=25
The zeros of 25(3-5x)^2-4=25 are x=(-sqrt(29)+15)/(25),x=(sqrt(29)+15)/(25)