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

The solutions to the equation (x)=-3(x-2)^2+4 are x=1,x= 8/3

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

The zeros of (x)=-3(x-2)^2+4 are x=1,x= 8/3

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

x = 1, x = \frac{8}{3}

Decimal Notation

x = 1, x = 2.66666 \dots

Solution steps

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

Expand

- 3 (x - 2)^{2} + 4 : - 3 x^{2} + 12 x - 8

x = - 3 x^{2} + 12 x - 8

Switch sides

- 3 x^{2} + 12 x - 8 = x

Move

x

to the left side

- 3 x^{2} + 11 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 ( - 3 ) ( - 8 )}{2 ( - 3 )}

1 1^{2} - 4 (- 3) (- 8) = 5

x_{1, 2} = \frac{- 11 \pm 5}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 11 + 5}{2 ( - 3 )}, x_{2} = \frac{- 11 - 5}{2 ( - 3 )}

x = \frac{- 11 + 5}{2 ( - 3 )} : 1

x = \frac{- 11 - 5}{2 ( - 3 )} : \frac{8}{3}

The solutions to the quadratic equation are:

x = 1, x = \frac{8}{3}

6 x^{2} - 24 x = 0

x^{2} = 8 x + 105

14 x^{2} - 4 = x

x, \frac{- x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

3 - x^{2} = 5 x^{2} - 9

What are the solutions to the equation (x)=-3(x-2)^2+4 ?
The solutions to the equation (x)=-3(x-2)^2+4 are x=1,x= 8/3
Find the zeros of (x)=-3(x-2)^2+4
The zeros of (x)=-3(x-2)^2+4 are x=1,x= 8/3