What are the solutions to the equation (3x+1)(x-4)=(2x+1)(x-1)-37 ?

The solutions to the equation (3x+1)(x-4)=(2x+1)(x-1)-37 are x=5+3i,x=5-3i

Find the zeros of (3x+1)(x-4)=(2x+1)(x-1)-37

The zeros of (3x+1)(x-4)=(2x+1)(x-1)-37 are x=5+3i,x=5-3i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(3 x + 1) (x - 4) = (2 x + 1) (x - 1) - 37

x = 5 + 3 i, x = 5 - 3 i

Solution steps

(3 x + 1) (x - 4) = (2 x + 1) (x - 1) - 37

Expand

(3 x + 1) (x - 4) : 3 x^{2} - 11 x - 4

Expand

(2 x + 1) (x - 1) - 37 : 2 x^{2} - x - 38

3 x^{2} - 11 x - 4 = 2 x^{2} - x - 38

Move

38

to the left side

3 x^{2} - 11 x + 34 = 2 x^{2} - x

Move

x

to the left side

3 x^{2} - 10 x + 34 = 2 x^{2}

Move

2 x^{2}

to the left side

x^{2} - 10 x + 34 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 34}{2 \cdot 1}

Simplify

(- 10)^{2} - 4 \cdot 1 \cdot 34 : 6 i

x_{1, 2} = \frac{- ( - 10 ) \pm 6 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 6 i}{2 \cdot 1}, x_{2} = \frac{- ( - 10 ) - 6 i}{2 \cdot 1}

x = \frac{- ( - 10 ) + 6 i}{2 \cdot 1} : 5 + 3 i

x = \frac{- ( - 10 ) - 6 i}{2 \cdot 1} : 5 - 3 i

The solutions to the quadratic equation are:

x = 5 + 3 i, x = 5 - 3 i

7 x^{2} - 567 = 0

- x^{2} - 2 + 3 = 0

x^{2} - 9 x - 26 = 0

y, - x^{2} y^{2} = k

x^{2} + 14 x - 15 = 0

What are the solutions to the equation (3x+1)(x-4)=(2x+1)(x-1)-37 ?
The solutions to the equation (3x+1)(x-4)=(2x+1)(x-1)-37 are x=5+3i,x=5-3i
Find the zeros of (3x+1)(x-4)=(2x+1)(x-1)-37
The zeros of (3x+1)(x-4)=(2x+1)(x-1)-37 are x=5+3i,x=5-3i