What are the solutions to the equation 3300=-x^2+160x-1500 ?

The solutions to the equation 3300=-x^2+160x-1500 are x=40,x=120

Find the zeros of 3300=-x^2+160x-1500

The zeros of 3300=-x^2+160x-1500 are x=40,x=120

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3300 = - x^{2} + 160 x - 1500

x = 40, x = 120

Solution steps

3300 = - x^{2} + 160 x - 1500

Switch sides

- x^{2} + 160 x - 1500 = 3300

Move

3300

to the left side

- x^{2} + 160 x - 4800 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 160 \pm 16 0 ^{2} - 4 ( - 1 ) ( - 4800 )}{2 ( - 1 )}

16 0^{2} - 4 (- 1) (- 4800) = 80

x_{1, 2} = \frac{- 160 \pm 80}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 160 + 80}{2 ( - 1 )}, x_{2} = \frac{- 160 - 80}{2 ( - 1 )}

x = \frac{- 160 + 80}{2 ( - 1 )} : 40

x = \frac{- 160 - 80}{2 ( - 1 )} : 120

The solutions to the quadratic equation are:

x = 40, x = 120

- 5 = - 3 a - 2 a

lo g_{125} (\frac{1}{25})

- \frac{11}{6} v \geq - 66

- 1 - (\frac{1}{7})

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

What are the solutions to the equation 3300=-x^2+160x-1500 ?
The solutions to the equation 3300=-x^2+160x-1500 are x=40,x=120
Find the zeros of 3300=-x^2+160x-1500
The zeros of 3300=-x^2+160x-1500 are x=40,x=120