What are the solutions to the equation (x+2)(x+5)=28 ?

The solutions to the equation (x+2)(x+5)=28 are x=2,x=-9

Find the zeros of (x+2)(x+5)=28

The zeros of (x+2)(x+5)=28 are x=2,x=-9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x + 2) (x + 5) = 28

x = 2, x = - 9

Solution steps

(x + 2) (x + 5) = 28

Expand

(x + 2) (x + 5) : x^{2} + 7 x + 10

x^{2} + 7 x + 10 = 28

Move

28

to the left side

x^{2} + 7 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 18) = 11

x_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 11}{2 \cdot 1}, x_{2} = \frac{- 7 - 11}{2 \cdot 1}

x = \frac{- 7 + 11}{2 \cdot 1} : 2

x = \frac{- 7 - 11}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

x = 2, x = - 9

15 f^{2} + 2 f - 45

\frac{8 k + 9}{18} - \frac{2 k + 1}{18}

∣ x - 7 ∣ = 27

\frac{13}{3} = \frac{u}{11}

4 x^{2} + 16 x + 7

What are the solutions to the equation (x+2)(x+5)=28 ?
The solutions to the equation (x+2)(x+5)=28 are x=2,x=-9
Find the zeros of (x+2)(x+5)=28
The zeros of (x+2)(x+5)=28 are x=2,x=-9