What are the solutions to the equation 0.12q^2+200=-10q+1000 ?

The solutions to the equation 0.12q^2+200=-10q+1000 are q=50,q=-400/3

Find the zeros of 0.12q^2+200=-10q+1000

The zeros of 0.12q^2+200=-10q+1000 are q=50,q=-400/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0.12 q^{2} + 200 = - 10 q + 1000

q = 50, q = - \frac{400}{3}

Decimal Notation

q = 50, q = - 133.33333 \dots

Solution steps

0.12 q^{2} + 200 = - 10 q + 1000

Multiply both sides by

100

12 q^{2} + 20000 = - 1000 q + 100000

Move

100000

to the left side

12 q^{2} - 80000 = - 1000 q

Move

1000 q

to the left side

12 q^{2} - 80000 + 1000 q = 0

Write in the standard form

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

12 q^{2} + 1000 q - 80000 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 1000 \pm 100 0 ^{2} - 4 \cdot 12 ( - 80000 )}{2 \cdot 12}

100 0^{2} - 4 \cdot 12 (- 80000) = 2200

q_{1, 2} = \frac{- 1000 \pm 2200}{2 \cdot 12}

Separate the solutions

q_{1} = \frac{- 1000 + 2200}{2 \cdot 12}, q_{2} = \frac{- 1000 - 2200}{2 \cdot 12}

q = \frac{- 1000 + 2200}{2 \cdot 12} : 50

q = \frac{- 1000 - 2200}{2 \cdot 12} : - \frac{400}{3}

The solutions to the quadratic equation are:

q = 50, q = - \frac{400}{3}

7 x^{2} - 21 x - 28 = 0

a, f (a + h) = 3 (a + h)^{2} + 5 (a + h) - 2

(x - 6)^{2} = 2 (x + 18)

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

z^{2} + 20 = 2 \cdot (z - 3)

What are the solutions to the equation 0.12q^2+200=-10q+1000 ?
The solutions to the equation 0.12q^2+200=-10q+1000 are q=50,q=-400/3
Find the zeros of 0.12q^2+200=-10q+1000
The zeros of 0.12q^2+200=-10q+1000 are q=50,q=-400/3