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

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

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

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

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

x = 2, x = - 4

Solution steps

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

Write in the standard form

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

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

x_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 6}{2 \cdot 1}, x_{2} = \frac{- 2 - 6}{2 \cdot 1}

x = \frac{- 2 + 6}{2 \cdot 1} : 2

x = \frac{- 2 - 6}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 2, x = - 4

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

4 x^{2} - 35 x + 24 = 0

ω, \frac{π}{4} \cdot 2 0^{2} \cdot 50 = \frac{π}{4} ω^{2} \cdot 60

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

28090 = 25000 (1 + \frac{r}{100})^{2}

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