What are the solutions to the equation -4(x-3)^2+1=0 ?

The solutions to the equation -4(x-3)^2+1=0 are x= 7/2 ,x= 5/2

Find the zeros of -4(x-3)^2+1=0

The zeros of -4(x-3)^2+1=0 are x= 7/2 ,x= 5/2

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- 4 (x - 3)^{2} + 1 = 0

x = \frac{7}{2}, x = \frac{5}{2}

Decimal Notation

x = 3.5, x = 2.5

Solution steps

- 4 (x - 3)^{2} + 1 = 0

Move

1

to the right side

- 4 (x - 3)^{2} = - 1

Divide both sides by

- 4

(x - 3)^{2} = \frac{1}{4}

For

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

the solutions are

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

Solve

x - 3 = \frac{1}{4} : x = \frac{7}{2}

Solve

x - 3 = - \frac{1}{4} : x = \frac{5}{2}

The solutions to the quadratic equation are:

x = \frac{7}{2}, x = \frac{5}{2}

5 r^{2} + \frac{366}{3} r + 2 = 0

c^{2} + 4 c + 4 = 0

12 n^{2} + 12 n + 9 = 0

x, 16 x^{2} + 9 y^{2} + 16 z^{2} - 32 x - 36 y + 36 = 0

3 X^{2} - 2 X + 4 = 0

What are the solutions to the equation -4(x-3)^2+1=0 ?
The solutions to the equation -4(x-3)^2+1=0 are x= 7/2 ,x= 5/2
Find the zeros of -4(x-3)^2+1=0
The zeros of -4(x-3)^2+1=0 are x= 7/2 ,x= 5/2