What are the solutions to the equation 9(5x+2)^2-42=7 ?

The solutions to the equation 9(5x+2)^2-42=7 are x= 1/15 ,x=-13/15

Find the zeros of 9(5x+2)^2-42=7

The zeros of 9(5x+2)^2-42=7 are x= 1/15 ,x=-13/15

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9 (5 x + 2)^{2} - 42 = 7

x = \frac{1}{15}, x = - \frac{13}{15}

Solution steps

9 (5 x + 2)^{2} - 42 = 7

Move

42

to the right side

9 (5 x + 2)^{2} = 49

Divide both sides by

9

(5 x + 2)^{2} = \frac{49}{9}

For

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

the solutions are

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

Solve

5 x + 2 = \frac{49}{9} : x = \frac{1}{15}

Solve

5 x + 2 = - \frac{49}{9} : x = - \frac{13}{15}

The solutions to the quadratic equation are:

x = \frac{1}{15}, x = - \frac{13}{15}

x, f = x^{2} y^{2} - x^{2} y^{3} + 13

3 x^{2} + 4 x - 12 = 0

x, x^{2} - 2 x = 0

x^{2} + 9 x - 18 = 0

12 x^{2} + 4 x = 0

What are the solutions to the equation 9(5x+2)^2-42=7 ?
The solutions to the equation 9(5x+2)^2-42=7 are x= 1/15 ,x=-13/15
Find the zeros of 9(5x+2)^2-42=7
The zeros of 9(5x+2)^2-42=7 are x= 1/15 ,x=-13/15