What are the solutions to the equation (-x^2-2x+8)=0 ?

The solutions to the equation (-x^2-2x+8)=0 are x=2,x=-4

Find the zeros of (-x^2-2x+8)=0

The zeros of (-x^2-2x+8)=0 are x=2,x=-4

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(- x^{2} - 2 x + 8) = 0

x = 2, x = - 4

Solution steps

(- x^{2} - 2 x + 8) = 0

Factor

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

- (x - 2) (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x + 4 = 0

Solve

x - 2 = 0 : x = 2

Solve

x + 4 = 0 : x = - 4

The solutions to the quadratic equation are:

x = 2, x = - 4

\frac{1}{3} \cdot 18

\frac{3}{5} (\frac{x - 1}{3} + 1) + x = \frac{3}{4} (x - \frac{2}{3})

\frac{1}{3} π \cdot 2. 7^{2} \cdot 5

7 y^{2} - 42 y = 0

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

What are the solutions to the equation (-x^2-2x+8)=0 ?
The solutions to the equation (-x^2-2x+8)=0 are x=2,x=-4
Find the zeros of (-x^2-2x+8)=0
The zeros of (-x^2-2x+8)=0 are x=2,x=-4