What are the solutions to the equation 3320=-2x^2+180x-280 ?

The solutions to the equation 3320=-2x^2+180x-280 are x=30,x=60

Find the zeros of 3320=-2x^2+180x-280

The zeros of 3320=-2x^2+180x-280 are x=30,x=60

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3320 = - 2 x^{2} + 180 x - 280

x = 30, x = 60

Solution steps

3320 = - 2 x^{2} + 180 x - 280

Switch sides

- 2 x^{2} + 180 x - 280 = 3320

Move

3320

to the left side

- 2 x^{2} + 180 x - 3600 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 180 \pm 18 0 ^{2} - 4 ( - 2 ) ( - 3600 )}{2 ( - 2 )}

18 0^{2} - 4 (- 2) (- 3600) = 60

x_{1, 2} = \frac{- 180 \pm 60}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 180 + 60}{2 ( - 2 )}, x_{2} = \frac{- 180 - 60}{2 ( - 2 )}

x = \frac{- 180 + 60}{2 ( - 2 )} : 30

x = \frac{- 180 - 60}{2 ( - 2 )} : 60

The solutions to the quadratic equation are:

x = 30, x = 60

3 m^{2} - (5 m + 1) (2 m - 3) = 3

x^{2} + x - 6 = 9

x^{2} + 144 = 169

(1 - x) (- 2 - x) - 4 = 0

100 - q^{2} = q^{2} + 4 q + 4

What are the solutions to the equation 3320=-2x^2+180x-280 ?
The solutions to the equation 3320=-2x^2+180x-280 are x=30,x=60
Find the zeros of 3320=-2x^2+180x-280
The zeros of 3320=-2x^2+180x-280 are x=30,x=60