What are the solutions to the equation 15000=-0.5x^2+40x-300 ?

The solutions to the equation 15000=-0.5x^2+40x-300 are x=40-10sqrt(290)i,x=40+10sqrt(290)i

Find the zeros of 15000=-0.5x^2+40x-300

The zeros of 15000=-0.5x^2+40x-300 are x=40-10sqrt(290)i,x=40+10sqrt(290)i

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15000 = - 0.5 x^{2} + 40 x - 300

x = 40 - 10290 i, x = 40 + 10290 i

Solution steps

15000 = - 0.5 x^{2} + 40 x - 300

Multiply both sides by

10

150000 = - 5 x^{2} + 400 x - 3000

Switch sides

- 5 x^{2} + 400 x - 3000 = 150000

Move

150000

to the left side

- 5 x^{2} + 400 x - 153000 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 400 \pm 40 0 ^{2} - 4 ( - 5 ) ( - 153000 )}{2 ( - 5 )}

Simplify

40 0^{2} - 4 (- 5) (- 153000) : 100290 i

x_{1, 2} = \frac{- 400 \pm 100 290 i}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- 400 + 100 290 i}{2 ( - 5 )}, x_{2} = \frac{- 400 - 100 290 i}{2 ( - 5 )}

x = \frac{- 400 + 100 290 i}{2 ( - 5 )} : 40 - 10290 i

x = \frac{- 400 - 100 290 i}{2 ( - 5 )} : 40 + 10290 i

The solutions to the quadratic equation are:

x = 40 - 10290 i, x = 40 + 10290 i

\frac{x ^{2}}{3} = \frac{11}{6} x + 1

(k + 2) x^{2} + 5 x - 3 = 0

(x + 2)^{2} - 36 = 0

t^{2} - 36 = 0

\frac{X ^{2}}{64} - \frac{X ^{2}}{9} = 1

What are the solutions to the equation 15000=-0.5x^2+40x-300 ?
The solutions to the equation 15000=-0.5x^2+40x-300 are x=40-10sqrt(290)i,x=40+10sqrt(290)i
Find the zeros of 15000=-0.5x^2+40x-300
The zeros of 15000=-0.5x^2+40x-300 are x=40-10sqrt(290)i,x=40+10sqrt(290)i