What are the solutions to the equation w^2+w-12=0 ?

The solutions to the equation w^2+w-12=0 are w=3,w=-4

Find the zeros of w^2+w-12=0

The zeros of w^2+w-12=0 are w=3,w=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

w^{2} + w - 12 = 0

w = 3, w = - 4

Solution steps

w^{2} + w - 12 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 12) = 7

w_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- 1 + 7}{2 \cdot 1}, w_{2} = \frac{- 1 - 7}{2 \cdot 1}

w = \frac{- 1 + 7}{2 \cdot 1} : 3

w = \frac{- 1 - 7}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

w = 3, w = - 4

2 f = - 12

3 (2 x + 4) = 6 x - 12

5 x^{2} - 48 x + 20 = 0

(2 x - 3) (3 x - 4) - (x - 13) (x - 4) = 40

What are the solutions to the equation w^2+w-12=0 ?
The solutions to the equation w^2+w-12=0 are w=3,w=-4
Find the zeros of w^2+w-12=0
The zeros of w^2+w-12=0 are w=3,w=-4