What are the solutions to the equation 180+6q-4q^2=2q^2 ?

The solutions to the equation 180+6q-4q^2=2q^2 are q=-5,q=6

Find the zeros of 180+6q-4q^2=2q^2

The zeros of 180+6q-4q^2=2q^2 are q=-5,q=6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

180 + 6 q - 4 q^{2} = 2 q^{2}

q = - 5, q = 6

Solution steps

180 + 6 q - 4 q^{2} = 2 q^{2}

Move

2 q^{2}

to the left side

180 + 6 q - 6 q^{2} = 0

Write in the standard form

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

- 6 q^{2} + 6 q + 180 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 6 ) \cdot 180}{2 ( - 6 )}

6^{2} - 4 (- 6) \cdot 180 = 66

q_{1, 2} = \frac{- 6 \pm 66}{2 ( - 6 )}

Separate the solutions

q_{1} = \frac{- 6 + 66}{2 ( - 6 )}, q_{2} = \frac{- 6 - 66}{2 ( - 6 )}

q = \frac{- 6 + 66}{2 ( - 6 )} : - 5

q = \frac{- 6 - 66}{2 ( - 6 )} : 6

The solutions to the quadratic equation are:

q = - 5, q = 6

24 x^{2} + 15 x = 0

x^{2} + 11 x - 4 = 0

x^{2} + 11 x + 28 = 70

- x^{2} + 4 x - 7

s^{2} - 3 s - 10 = 0

What are the solutions to the equation 180+6q-4q^2=2q^2 ?
The solutions to the equation 180+6q-4q^2=2q^2 are q=-5,q=6
Find the zeros of 180+6q-4q^2=2q^2
The zeros of 180+6q-4q^2=2q^2 are q=-5,q=6