What are the solutions to the equation 8+4x-x^2=0 ?

The solutions to the equation 8+4x-x^2=0 are x=-2(sqrt(3)-1),x=2(1+sqrt(3))

Find the zeros of 8+4x-x^2=0

The zeros of 8+4x-x^2=0 are x=-2(sqrt(3)-1),x=2(1+sqrt(3))

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8 + 4 x - x^{2} = 0

x = - 2 (3 - 1), x = 2 (1 + 3)

Decimal Notation

x = - 1.46410 \dots, x = 5.46410 \dots

Solution steps

8 + 4 x - x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- x^{2} + 4 x + 8 = 0

Solve with the quadratic formula

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

4^{2} - 4 (- 1) \cdot 8 = 43

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

Separate the solutions

x_{1} = \frac{- 4 + 4 3}{2 ( - 1 )}, x_{2} = \frac{- 4 - 4 3}{2 ( - 1 )}

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

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

The solutions to the quadratic equation are:

x = - 2 (3 - 1), x = 2 (1 + 3)

2 x (x - 3) = 1

5 (p^{2} - 4) - 1 = 4

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

3 - 2 x^{2} = - 5

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

What are the solutions to the equation 8+4x-x^2=0 ?
The solutions to the equation 8+4x-x^2=0 are x=-2(sqrt(3)-1),x=2(1+sqrt(3))
Find the zeros of 8+4x-x^2=0
The zeros of 8+4x-x^2=0 are x=-2(sqrt(3)-1),x=2(1+sqrt(3))