What are the solutions to the equation 2x^2-(x+2)(x-3)=12 ?

The solutions to the equation 2x^2-(x+2)(x-3)=12 are x=2,x=-3

Find the zeros of 2x^2-(x+2)(x-3)=12

The zeros of 2x^2-(x+2)(x-3)=12 are x=2,x=-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{2} - (x + 2) (x - 3) = 12

x = 2, x = - 3

Solution steps

2 x^{2} - (x + 2) (x - 3) = 12

Expand

2 x^{2} - (x + 2) (x - 3) : x^{2} + x + 6

x^{2} + x + 6 = 12

Move

12

to the left side

x^{2} + x - 6 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 6) = 5

x_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 5}{2 \cdot 1}, x_{2} = \frac{- 1 - 5}{2 \cdot 1}

x = \frac{- 1 + 5}{2 \cdot 1} : 2

x = \frac{- 1 - 5}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 2, x = - 3

x^{2} + 8 x = 6 x + 3

x^{2} - 2 x - 3 = x - 3

7 p^{2} - 7 p = 7 p

(x + 1)^{2} = 14

(x + 20) (x - 5) = 350

What are the solutions to the equation 2x^2-(x+2)(x-3)=12 ?
The solutions to the equation 2x^2-(x+2)(x-3)=12 are x=2,x=-3
Find the zeros of 2x^2-(x+2)(x-3)=12
The zeros of 2x^2-(x+2)(x-3)=12 are x=2,x=-3