What are the solutions to the equation 1500-20x-2x^2=0 ?

The solutions to the equation 1500-20x-2x^2=0 are x=-5(1+sqrt(31)),x=5(sqrt(31)-1)

Find the zeros of 1500-20x-2x^2=0

The zeros of 1500-20x-2x^2=0 are x=-5(1+sqrt(31)),x=5(sqrt(31)-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1500 - 20 x - 2 x^{2} = 0

x = - 5 (1 + 31), x = 5 (31 - 1)

Decimal Notation

x = - 32.83882 \dots, x = 22.83882 \dots

Solution steps

1500 - 20 x - 2 x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 x^{2} - 20 x + 1500 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 ( - 2 ) \cdot 1500}{2 ( - 2 )}

(- 20)^{2} - 4 (- 2) \cdot 1500 = 2031

x_{1, 2} = \frac{- ( - 20 ) \pm 20 31}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 20 31}{2 ( - 2 )}, x_{2} = \frac{- ( - 20 ) - 20 31}{2 ( - 2 )}

x = \frac{- ( - 20 ) + 20 31}{2 ( - 2 )} : - 5 (1 + 31)

x = \frac{- ( - 20 ) - 20 31}{2 ( - 2 )} : 5 (31 - 1)

The solutions to the quadratic equation are:

x = - 5 (1 + 31), x = 5 (31 - 1)

q^{2} + 6 q + 9 = 0

y, x = y^{2} + 2 b y - b^{2}

x (x + 10) = 144

x, 6 x^{2} - y^{2} = 0

x (x - 5) + \frac{3}{2} = 2

What are the solutions to the equation 1500-20x-2x^2=0 ?
The solutions to the equation 1500-20x-2x^2=0 are x=-5(1+sqrt(31)),x=5(sqrt(31)-1)
Find the zeros of 1500-20x-2x^2=0
The zeros of 1500-20x-2x^2=0 are x=-5(1+sqrt(31)),x=5(sqrt(31)-1)