What are the solutions to the equation x^2+20x-800=0 ?

The solutions to the equation x^2+20x-800=0 are x=20,x=-40

Find the zeros of x^2+20x-800=0

The zeros of x^2+20x-800=0 are x=20,x=-40

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x^{2} + 20 x - 800 = 0

x = 20, x = - 40

Solution steps

x^{2} + 20 x - 800 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot ( - 800 )}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot (- 800) = 60

x_{1, 2} = \frac{- 20 \pm 60}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 20 + 60}{2 \cdot 1}, x_{2} = \frac{- 20 - 60}{2 \cdot 1}

x = \frac{- 20 + 60}{2 \cdot 1} : 20

x = \frac{- 20 - 60}{2 \cdot 1} : - 40

The solutions to the quadratic equation are:

x = 20, x = - 40

\frac{10}{7 - x} \geq x

u + 8 \geq 10

- 2 \cdot 10

\frac{- 6}{2 ( - 3 )}

2 - 2 n = 3 n + 17

What are the solutions to the equation x^2+20x-800=0 ?
The solutions to the equation x^2+20x-800=0 are x=20,x=-40
Find the zeros of x^2+20x-800=0
The zeros of x^2+20x-800=0 are x=20,x=-40