What are the solutions to the equation (x-14)(x+10)=-104 ?

The solutions to the equation (x-14)(x+10)=-104 are x=2(1+sqrt(10)),x=2(1-sqrt(10))

Find the zeros of (x-14)(x+10)=-104

The zeros of (x-14)(x+10)=-104 are x=2(1+sqrt(10)),x=2(1-sqrt(10))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x - 14) (x + 10) = - 104

x = 2 (1 + 10), x = 2 (1 - 10)

Decimal Notation

x = 8.32455 \dots, x = - 4.32455 \dots

Solution steps

(x - 14) (x + 10) = - 104

Expand

(x - 14) (x + 10) : x^{2} - 4 x - 140

x^{2} - 4 x - 140 = - 104

Move

104

to the left side

x^{2} - 4 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 36 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 36) = 410

x_{1, 2} = \frac{- ( - 4 ) \pm 4 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 10}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 4 10}{2 \cdot 1}

x = \frac{- ( - 4 ) + 4 10}{2 \cdot 1} : 2 (1 + 10)

x = \frac{- ( - 4 ) - 4 10}{2 \cdot 1} : 2 (1 - 10)

The solutions to the quadratic equation are:

x = 2 (1 + 10), x = 2 (1 - 10)

8 \cdot 32

- 13 < 3 x + 8 \leq 23

10 x - 4 x + 6 < 3 (x + 5)

5 x^{2} - 245

\frac{x}{2} + 3 = 8

What are the solutions to the equation (x-14)(x+10)=-104 ?
The solutions to the equation (x-14)(x+10)=-104 are x=2(1+sqrt(10)),x=2(1-sqrt(10))
Find the zeros of (x-14)(x+10)=-104
The zeros of (x-14)(x+10)=-104 are x=2(1+sqrt(10)),x=2(1-sqrt(10))