What are the solutions to the equation 820(1+x)^2=2000+2000x ?

The solutions to the equation 820(1+x)^2=2000+2000x are x= 59/41 ,x=-1

Find the zeros of 820(1+x)^2=2000+2000x

The zeros of 820(1+x)^2=2000+2000x are x= 59/41 ,x=-1

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820 (1 + x)^{2} = 2000 + 2000 x

x = \frac{59}{41}, x = - 1

Solution steps

820 (1 + x)^{2} = 2000 + 2000 x

Expand

820 (1 + x)^{2} : 820 + 1640 x + 820 x^{2}

820 + 1640 x + 820 x^{2} = 2000 + 2000 x

Move

2000 x

to the left side

820 x^{2} - 360 x + 820 = 2000

Move

2000

to the left side

820 x^{2} - 360 x - 1180 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 360 ) \pm ( - 360 ) ^{2} - 4 \cdot 820 ( - 1180 )}{2 \cdot 820}

(- 360)^{2} - 4 \cdot 820 (- 1180) = 2000

x_{1, 2} = \frac{- ( - 360 ) \pm 2000}{2 \cdot 820}

Separate the solutions

x_{1} = \frac{- ( - 360 ) + 2000}{2 \cdot 820}, x_{2} = \frac{- ( - 360 ) - 2000}{2 \cdot 820}

x = \frac{- ( - 360 ) + 2000}{2 \cdot 820} : \frac{59}{41}

x = \frac{- ( - 360 ) - 2000}{2 \cdot 820} : - 1

The solutions to the quadratic equation are:

x = \frac{59}{41}, x = - 1

4 m^{2} - 7 m - 2 = 0

5 \cdot \frac{3}{2}

\frac{x}{∣ x + 4 ∣} < 0

x^{2} < 81

8 - x^{3}

What are the solutions to the equation 820(1+x)^2=2000+2000x ?
The solutions to the equation 820(1+x)^2=2000+2000x are x= 59/41 ,x=-1
Find the zeros of 820(1+x)^2=2000+2000x
The zeros of 820(1+x)^2=2000+2000x are x= 59/41 ,x=-1