What are the solutions to the equation (40-x)(20+2x)=1200 ?

The solutions to the equation (40-x)(20+2x)=1200 are x=10,x=20

Find the zeros of (40-x)(20+2x)=1200

The zeros of (40-x)(20+2x)=1200 are x=10,x=20

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(40 - x) (20 + 2 x) = 1200

x = 10, x = 20

Solution steps

(40 - x) (20 + 2 x) = 1200

Expand

(40 - x) (20 + 2 x) : 800 + 60 x - 2 x^{2}

800 + 60 x - 2 x^{2} = 1200

Move

1200

to the left side

- 2 x^{2} + 60 x - 400 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 60 \pm 6 0 ^{2} - 4 ( - 2 ) ( - 400 )}{2 ( - 2 )}

6 0^{2} - 4 (- 2) (- 400) = 20

x_{1, 2} = \frac{- 60 \pm 20}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 60 + 20}{2 ( - 2 )}, x_{2} = \frac{- 60 - 20}{2 ( - 2 )}

x = \frac{- 60 + 20}{2 ( - 2 )} : 10

x = \frac{- 60 - 20}{2 ( - 2 )} : 20

The solutions to the quadratic equation are:

x = 10, x = 20

x (x + 1) - 19 = 2 x + 1

1 x^{2} + 0 x - 4 = 0

6 4^{2} = 2 x^{2}

(a - 2) (a - 4) = 0

11.13 X^{2} + 18.116 X - 16 = 0

What are the solutions to the equation (40-x)(20+2x)=1200 ?
The solutions to the equation (40-x)(20+2x)=1200 are x=10,x=20
Find the zeros of (40-x)(20+2x)=1200
The zeros of (40-x)(20+2x)=1200 are x=10,x=20