What are the solutions to the equation x^2+41x=-40 ?

The solutions to the equation x^2+41x=-40 are x=-1,x=-40

Find the zeros of x^2+41x=-40

The zeros of x^2+41x=-40 are x=-1,x=-40

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x^{2} + 41 x = - 40

x = - 1, x = - 40

Solution steps

x^{2} + 41 x = - 40

Move

40

to the left side

x^{2} + 41 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 41 \pm 4 1 ^{2} - 4 \cdot 1 \cdot 40}{2 \cdot 1}

4 1^{2} - 4 \cdot 1 \cdot 40 = 39

x_{1, 2} = \frac{- 41 \pm 39}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 41 + 39}{2 \cdot 1}, x_{2} = \frac{- 41 - 39}{2 \cdot 1}

x = \frac{- 41 + 39}{2 \cdot 1} : - 1

x = \frac{- 41 - 39}{2 \cdot 1} : - 40

The solutions to the quadratic equation are:

x = - 1, x = - 40

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4 (x - 3)^{2} + 24 x = 0

A (n) = (700 + n) (10 - 0.001 n)

- 5 x^{2} + 350 x - 3000 = 0

3 (x - 9)^{2} = 21

What are the solutions to the equation x^2+41x=-40 ?
The solutions to the equation x^2+41x=-40 are x=-1,x=-40
Find the zeros of x^2+41x=-40
The zeros of x^2+41x=-40 are x=-1,x=-40