What are the solutions to the equation -x^2+40x-1200=0 ?

The solutions to the equation -x^2+40x-1200=0 are x=20-20sqrt(2)i,x=20+20sqrt(2)i

Find the zeros of -x^2+40x-1200=0

The zeros of -x^2+40x-1200=0 are x=20-20sqrt(2)i,x=20+20sqrt(2)i

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- x^{2} + 40 x - 1200 = 0

x = 20 - 202 i, x = 20 + 202 i

Solution steps

- x^{2} + 40 x - 1200 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 1 ) ( - 1200 )}{2 ( - 1 )}

Simplify

4 0^{2} - 4 (- 1) (- 1200) : 402 i

x_{1, 2} = \frac{- 40 \pm 40 2 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 40 + 40 2 i}{2 ( - 1 )}, x_{2} = \frac{- 40 - 40 2 i}{2 ( - 1 )}

x = \frac{- 40 + 40 2 i}{2 ( - 1 )} : 20 - 202 i

x = \frac{- 40 - 40 2 i}{2 ( - 1 )} : 20 + 202 i

The solutions to the quadratic equation are:

x = 20 - 202 i, x = 20 + 202 i

\frac{1}{7} x^{2} - 3 = 4

x, y = x^{2} - 20 x

3 x^{2} + 14 x - 1 = 7 x - 3

3 x^{2} + 10 x + 20 = 10 - 3 x

x, f (x + 2) = 4 x^{2} + 3

What are the solutions to the equation -x^2+40x-1200=0 ?
The solutions to the equation -x^2+40x-1200=0 are x=20-20sqrt(2)i,x=20+20sqrt(2)i
Find the zeros of -x^2+40x-1200=0
The zeros of -x^2+40x-1200=0 are x=20-20sqrt(2)i,x=20+20sqrt(2)i