What are the solutions to the equation (x+8)(x+4)=5^1 ?

The solutions to the equation (x+8)(x+4)=5^1 are x=-3,x=-9

Find the zeros of (x+8)(x+4)=5^1

The zeros of (x+8)(x+4)=5^1 are x=-3,x=-9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x + 8) (x + 4) = 5^{1}

x = - 3, x = - 9

Solution steps

(x + 8) (x + 4) = 5^{1}

Expand

(x + 8) (x + 4) : x^{2} + 12 x + 32

Expand

5^{1} : 5

x^{2} + 12 x + 32 = 5

Move

5

to the left side

x^{2} + 12 x + 27 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 27}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 27 = 6

x_{1, 2} = \frac{- 12 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 6}{2 \cdot 1}, x_{2} = \frac{- 12 - 6}{2 \cdot 1}

x = \frac{- 12 + 6}{2 \cdot 1} : - 3

x = \frac{- 12 - 6}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

x = - 3, x = - 9

6 (x + 1) = (4 - x) (7 + x)

0 = 12 x^{2} - 180 x + 464

x, f = x^{2} + y^{2} - z^{2}

5 x^{2} + 2 = - 4 x^{2}

s^{2} + 10 s + 26 = 0

What are the solutions to the equation (x+8)(x+4)=5^1 ?
The solutions to the equation (x+8)(x+4)=5^1 are x=-3,x=-9
Find the zeros of (x+8)(x+4)=5^1
The zeros of (x+8)(x+4)=5^1 are x=-3,x=-9