What are the solutions to the equation (x-2)(x+3)=14 ?

The solutions to the equation (x-2)(x+3)=14 are x=4,x=-5

Find the zeros of (x-2)(x+3)=14

The zeros of (x-2)(x+3)=14 are x=4,x=-5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x - 2) (x + 3) = 14

x = 4, x = - 5

Solution steps

(x - 2) (x + 3) = 14

Expand

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

x^{2} + x - 6 = 14

Move

14

to the left side

x^{2} + x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 20) = 9

x_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 9}{2 \cdot 1}, x_{2} = \frac{- 1 - 9}{2 \cdot 1}

x = \frac{- 1 + 9}{2 \cdot 1} : 4

x = \frac{- 1 - 9}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 4, x = - 5

\frac{16}{5}

20 x^{2} - 5 x^{6}

\frac{77}{6} - (\frac{11}{16})^{2}

∣ 5 + 3 i ∣

x^{3} - 2 x^{2} < x^{2} (x - 2) - 3 (x - 1)

What are the solutions to the equation (x-2)(x+3)=14 ?
The solutions to the equation (x-2)(x+3)=14 are x=4,x=-5
Find the zeros of (x-2)(x+3)=14
The zeros of (x-2)(x+3)=14 are x=4,x=-5