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

The solutions to the equation x-3=-(x-4)^2+5 are x=(7-sqrt(17))/2 ,x=(7+sqrt(17))/2

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

The zeros of x-3=-(x-4)^2+5 are x=(7-sqrt(17))/2 ,x=(7+sqrt(17))/2

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

x = \frac{7 - 17}{2}, x = \frac{7 + 17}{2}

Solution steps

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

Expand

- (x - 4)^{2} + 5 : - x^{2} + 8 x - 11

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

Switch sides

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

Move

3

to the left side

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

Move

x

to the left side

- x^{2} + 7 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 1 ) ( - 8 )}{2 ( - 1 )}

7^{2} - 4 (- 1) (- 8) = 17

x_{1, 2} = \frac{- 7 \pm 17}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 7 + 17}{2 ( - 1 )}, x_{2} = \frac{- 7 - 17}{2 ( - 1 )}

x = \frac{- 7 + 17}{2 ( - 1 )} : \frac{7 - 17}{2}

x = \frac{- 7 - 17}{2 ( - 1 )} : \frac{7 + 17}{2}

The solutions to the quadratic equation are:

x = \frac{7 - 17}{2}, x = \frac{7 + 17}{2}

0.0004

\frac{\frac{3 !}{2 ! 1 !}}{3 6 ^{2}}

2 l + 2 w

8 \geq 11 + y

x^{4} \cdot x^{\frac{1}{2}}

What are the solutions to the equation x-3=-(x-4)^2+5 ?
The solutions to the equation x-3=-(x-4)^2+5 are x=(7-sqrt(17))/2 ,x=(7+sqrt(17))/2
Find the zeros of x-3=-(x-4)^2+5
The zeros of x-3=-(x-4)^2+5 are x=(7-sqrt(17))/2 ,x=(7+sqrt(17))/2