What are the solutions to the equation 4000=-5x^2+400x-2550 ?

The solutions to the equation 4000=-5x^2+400x-2550 are x=40-sqrt(290),x=40+sqrt(290)

Find the zeros of 4000=-5x^2+400x-2550

The zeros of 4000=-5x^2+400x-2550 are x=40-sqrt(290),x=40+sqrt(290)

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4000 = - 5 x^{2} + 400 x - 2550

x = 40 - 290, x = 40 + 290

Decimal

x = 22.97061 \dots, x = 57.02938 \dots

Solution steps

4000 = - 5 x^{2} + 400 x - 2550

Switch sides

- 5 x^{2} + 400 x - 2550 = 4000

Move

4000

to the left side

- 5 x^{2} + 400 x - 6550 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 400 \pm 40 0 ^{2} - 4 ( - 5 ) ( - 6550 )}{2 ( - 5 )}

40 0^{2} - 4 (- 5) (- 6550) = 10290

x_{1, 2} = \frac{- 400 \pm 10 290}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- 400 + 10 290}{2 ( - 5 )}, x_{2} = \frac{- 400 - 10 290}{2 ( - 5 )}

x = \frac{- 400 + 10 290}{2 ( - 5 )} : 40 - 290

x = \frac{- 400 - 10 290}{2 ( - 5 )} : 40 + 290

The solutions to the quadratic equation are:

x = 40 - 290, x = 40 + 290

x^{2} + 2 x - 12 = 0

x^{2} + 10 x + 23 = 0

u^{2} - 6 u + 8 = 0

- x^{2} + 4 x + 10 = 0

- 8 x + 96 - 4 x^{2} = 0

What are the solutions to the equation 4000=-5x^2+400x-2550 ?
The solutions to the equation 4000=-5x^2+400x-2550 are x=40-sqrt(290),x=40+sqrt(290)
Find the zeros of 4000=-5x^2+400x-2550
The zeros of 4000=-5x^2+400x-2550 are x=40-sqrt(290),x=40+sqrt(290)