What are the solutions to the equation 9(2x-3)^2+7=448 ?

The solutions to the equation 9(2x-3)^2+7=448 are x=5,x=-2

Find the zeros of 9(2x-3)^2+7=448

The zeros of 9(2x-3)^2+7=448 are x=5,x=-2

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9 (2 x - 3)^{2} + 7 = 448

x = 5, x = - 2

Solution steps

9 (2 x - 3)^{2} + 7 = 448

Move

7

to the right side

9 (2 x - 3)^{2} = 441

Divide both sides by

9

(2 x - 3)^{2} = 49

For

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

the solutions are

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

Solve

2 x - 3 = 49 : x = 5

Solve

2 x - 3 = - 49 : x = - 2

The solutions to the quadratic equation are:

x = 5, x = - 2

4 x - (9 - 3 x) = 8 x - 1

2 k^{2} - k - 6

- 4 (x + 4) = 2 (x + 10)

9^{2} + x^{2} = 1 2^{2}

\frac{30}{4.9} = x^{2}

What are the solutions to the equation 9(2x-3)^2+7=448 ?
The solutions to the equation 9(2x-3)^2+7=448 are x=5,x=-2
Find the zeros of 9(2x-3)^2+7=448
The zeros of 9(2x-3)^2+7=448 are x=5,x=-2