What are the solutions to the equation 2500=-0.5x^2+40x-300 ?

The solutions to the equation 2500=-0.5x^2+40x-300 are x=40-20sqrt(10)i,x=40+20sqrt(10)i

Find the zeros of 2500=-0.5x^2+40x-300

The zeros of 2500=-0.5x^2+40x-300 are x=40-20sqrt(10)i,x=40+20sqrt(10)i

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2500 = - 0.5 x^{2} + 40 x - 300

x = 40 - 2010 i, x = 40 + 2010 i

Solution steps

2500 = - 0.5 x^{2} + 40 x - 300

Multiply both sides by

10

25000 = - 5 x^{2} + 400 x - 3000

Switch sides

- 5 x^{2} + 400 x - 3000 = 25000

Move

25000

to the left side

- 5 x^{2} + 400 x - 28000 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 400 \pm 40 0 ^{2} - 4 ( - 5 ) ( - 28000 )}{2 ( - 5 )}

Simplify

40 0^{2} - 4 (- 5) (- 28000) : 20010 i

x_{1, 2} = \frac{- 400 \pm 200 10 i}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- 400 + 200 10 i}{2 ( - 5 )}, x_{2} = \frac{- 400 - 200 10 i}{2 ( - 5 )}

x = \frac{- 400 + 200 10 i}{2 ( - 5 )} : 40 - 2010 i

x = \frac{- 400 - 200 10 i}{2 ( - 5 )} : 40 + 2010 i

The solutions to the quadratic equation are:

x = 40 - 2010 i, x = 40 + 2010 i

x, (x + y)^{2} = 1

u^{2} - 4 u + 1 = 0

- 18 x^{2} + 18 x = 0

(x + 1)^{2} = 56

y^{2} - 6 y - 15 = 0

What are the solutions to the equation 2500=-0.5x^2+40x-300 ?
The solutions to the equation 2500=-0.5x^2+40x-300 are x=40-20sqrt(10)i,x=40+20sqrt(10)i
Find the zeros of 2500=-0.5x^2+40x-300
The zeros of 2500=-0.5x^2+40x-300 are x=40-20sqrt(10)i,x=40+20sqrt(10)i