What are the solutions to the equation 50=-0.5x^2+40x-300 ?

The solutions to the equation 50=-0.5x^2+40x-300 are x=10,x=70

Find the zeros of 50=-0.5x^2+40x-300

The zeros of 50=-0.5x^2+40x-300 are x=10,x=70

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50 = - 0.5 x^{2} + 40 x - 300

x = 10, x = 70

Solution steps

50 = - 0.5 x^{2} + 40 x - 300

Multiply both sides by

10

500 = - 5 x^{2} + 400 x - 3000

Switch sides

- 5 x^{2} + 400 x - 3000 = 500

Move

500

to the left side

- 5 x^{2} + 400 x - 3500 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 400 \pm 40 0 ^{2} - 4 ( - 5 ) ( - 3500 )}{2 ( - 5 )}

40 0^{2} - 4 (- 5) (- 3500) = 300

x_{1, 2} = \frac{- 400 \pm 300}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- 400 + 300}{2 ( - 5 )}, x_{2} = \frac{- 400 - 300}{2 ( - 5 )}

x = \frac{- 400 + 300}{2 ( - 5 )} : 10

x = \frac{- 400 - 300}{2 ( - 5 )} : 70

The solutions to the quadratic equation are:

x = 10, x = 70

2 x^{2} - 3 x = 2 x + 15

n (x) = - 2 x^{2}

2 x + x^{2} = - 13

6 x^{2} - 16 x - 6 = 0

10 x^{2} + 25 x = 15

What are the solutions to the equation 50=-0.5x^2+40x-300 ?
The solutions to the equation 50=-0.5x^2+40x-300 are x=10,x=70
Find the zeros of 50=-0.5x^2+40x-300
The zeros of 50=-0.5x^2+40x-300 are x=10,x=70