What are the solutions to the equation -0.1q^2+1800=20q+550 ?

The solutions to the equation -0.1q^2+1800=20q+550 are q=-250,q=50

Find the zeros of -0.1q^2+1800=20q+550

The zeros of -0.1q^2+1800=20q+550 are q=-250,q=50

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- 0.1 q^{2} + 1800 = 20 q + 550

q = - 250, q = 50

Solution steps

- 0.1 q^{2} + 1800 = 20 q + 550

Multiply both sides by

10

- q^{2} + 18000 = 200 q + 5500

Move

5500

to the left side

- q^{2} + 12500 = 200 q

Move

200 q

to the left side

- q^{2} + 12500 - 200 q = 0

Write in the standard form

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

- q^{2} - 200 q + 12500 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 200 ) \pm ( - 200 ) ^{2} - 4 ( - 1 ) \cdot 12500}{2 ( - 1 )}

(- 200)^{2} - 4 (- 1) \cdot 12500 = 300

q_{1, 2} = \frac{- ( - 200 ) \pm 300}{2 ( - 1 )}

Separate the solutions

q_{1} = \frac{- ( - 200 ) + 300}{2 ( - 1 )}, q_{2} = \frac{- ( - 200 ) - 300}{2 ( - 1 )}

q = \frac{- ( - 200 ) + 300}{2 ( - 1 )} : - 250

q = \frac{- ( - 200 ) - 300}{2 ( - 1 )} : 50

The solutions to the quadratic equation are:

q = - 250, q = 50

n^{2} + 2 n + 4 = 3

x, y x - 3 y + x^{2} + 3 x - 15 = 0

2 x^{2} = 12 x - 18

x^{2} = 49 x

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

What are the solutions to the equation -0.1q^2+1800=20q+550 ?
The solutions to the equation -0.1q^2+1800=20q+550 are q=-250,q=50
Find the zeros of -0.1q^2+1800=20q+550
The zeros of -0.1q^2+1800=20q+550 are q=-250,q=50