What are the solutions to the equation (m+3)(5m+1)=3(m-4)+7 ?

The solutions to the equation (m+3)(5m+1)=3(m-4)+7 are m=-1,m=-8/5

Find the zeros of (m+3)(5m+1)=3(m-4)+7

The zeros of (m+3)(5m+1)=3(m-4)+7 are m=-1,m=-8/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(m + 3) (5 m + 1) = 3 (m - 4) + 7

m = - 1, m = - \frac{8}{5}

Decimal Notation

m = - 1, m = - 1.6

Solution steps

(m + 3) (5 m + 1) = 3 (m - 4) + 7

Expand

(m + 3) (5 m + 1) : 5 m^{2} + 16 m + 3

Expand

3 (m - 4) + 7 : 3 m - 5

5 m^{2} + 16 m + 3 = 3 m - 5

Move

5

to the left side

5 m^{2} + 16 m + 8 = 3 m

Move

3 m

to the left side

5 m^{2} + 13 m + 8 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 5 \cdot 8}{2 \cdot 5}

1 3^{2} - 4 \cdot 5 \cdot 8 = 3

m_{1, 2} = \frac{- 13 \pm 3}{2 \cdot 5}

Separate the solutions

m_{1} = \frac{- 13 + 3}{2 \cdot 5}, m_{2} = \frac{- 13 - 3}{2 \cdot 5}

m = \frac{- 13 + 3}{2 \cdot 5} : - 1

m = \frac{- 13 - 3}{2 \cdot 5} : - \frac{8}{5}

The solutions to the quadratic equation are:

m = - 1, m = - \frac{8}{5}

0.0000987 x^{2} - 0.009 x + 0.0613 = 0

6^{4} lo g_{6} (3)

\frac{9 !}{5 !}

5 (\frac{2}{3})

x + 9 = - 4 x + 39

What are the solutions to the equation (m+3)(5m+1)=3(m-4)+7 ?
The solutions to the equation (m+3)(5m+1)=3(m-4)+7 are m=-1,m=-8/5
Find the zeros of (m+3)(5m+1)=3(m-4)+7
The zeros of (m+3)(5m+1)=3(m-4)+7 are m=-1,m=-8/5