What are the solutions to the equation (80-2x)(1200+200x)-1600x-6400=122000 ?

The solutions to the equation (80-2x)(1200+200x)-1600x-6400=122000 are x=3,x=27

Find the zeros of (80-2x)(1200+200x)-1600x-6400=122000

The zeros of (80-2x)(1200+200x)-1600x-6400=122000 are x=3,x=27

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(80 - 2 x) (1200 + 200 x) - 1600 x - 6400 = 122000

x = 3, x = 27

Solution steps

(80 - 2 x) (1200 + 200 x) - 1600 x - 6400 = 122000

Expand

(80 - 2 x) (1200 + 200 x) - 1600 x - 6400 : - 400 x^{2} + 12000 x + 89600

- 400 x^{2} + 12000 x + 89600 = 122000

Move

122000

to the left side

- 400 x^{2} + 12000 x - 32400 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12000 \pm 1200 0 ^{2} - 4 ( - 400 ) ( - 32400 )}{2 ( - 400 )}

1200 0^{2} - 4 (- 400) (- 32400) = 9600

x_{1, 2} = \frac{- 12000 \pm 9600}{2 ( - 400 )}

Separate the solutions

x_{1} = \frac{- 12000 + 9600}{2 ( - 400 )}, x_{2} = \frac{- 12000 - 9600}{2 ( - 400 )}

x = \frac{- 12000 + 9600}{2 ( - 400 )} : 3

x = \frac{- 12000 - 9600}{2 ( - 400 )} : 27

The solutions to the quadratic equation are:

x = 3, x = 27

λ^{2} - 7 λ + 10 = 0

(11 x - 18) (6 x + 10) = 0

3 x^{2} - 40 x + 20000 = 0

y, x = y^{2} - 4 y + 3

(8 x + 9)^{2} = 4

What are the solutions to the equation (80-2x)(1200+200x)-1600x-6400=122000 ?
The solutions to the equation (80-2x)(1200+200x)-1600x-6400=122000 are x=3,x=27
Find the zeros of (80-2x)(1200+200x)-1600x-6400=122000
The zeros of (80-2x)(1200+200x)-1600x-6400=122000 are x=3,x=27