What are the solutions to the equation 1653.75=1500(1+2x+x^2) ?

The solutions to the equation 1653.75=1500(1+2x+x^2) are x= 1/20 ,x=-41/20

Find the zeros of 1653.75=1500(1+2x+x^2)

The zeros of 1653.75=1500(1+2x+x^2) are x= 1/20 ,x=-41/20

1653.75 = 1500 (1 + 2 x + x^{2})

x = \frac{1}{20}, x = - \frac{41}{20}

Decimal

x = 0.05, x = - 2.05

Solution steps

1653.75 = 1500 (1 + 2 x + x^{2})

Multiply both sides by

100

165375 = 150000 (1 + 2 x + x^{2})

Expand

150000 (1 + 2 x + x^{2}) : 150000 + 300000 x + 150000 x^{2}

165375 = 150000 + 300000 x + 150000 x^{2}

Switch sides

150000 + 300000 x + 150000 x^{2} = 165375

Move

165375

to the left side

150000 x^{2} + 300000 x - 15375 = 0

Factor

150000 x^{2} + 300000 x - 15375 : 375 (20 x - 1) (20 x + 41)

375 (20 x - 1) (20 x + 41) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

20 x - 1 = 0 or 20 x + 41 = 0

Solve

20 x - 1 = 0 : x = \frac{1}{20}

Solve

20 x + 41 = 0 : x = - \frac{41}{20}

The solutions to the quadratic equation are:

x = \frac{1}{20}, x = - \frac{41}{20}

What are the solutions to the equation 1653.75=1500(1+2x+x^2) ?
The solutions to the equation 1653.75=1500(1+2x+x^2) are x= 1/20 ,x=-41/20
Find the zeros of 1653.75=1500(1+2x+x^2)
The zeros of 1653.75=1500(1+2x+x^2) are x= 1/20 ,x=-41/20