What are the solutions to the equation (2x-3)(3x-4)-(x-13)(x-4)=40 ?

The solutions to the equation (2x-3)(3x-4)-(x-13)(x-4)=40 are x=4,x=-4

Find the zeros of (2x-3)(3x-4)-(x-13)(x-4)=40

The zeros of (2x-3)(3x-4)-(x-13)(x-4)=40 are x=4,x=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2 x - 3) (3 x - 4) - (x - 13) (x - 4) = 40

x = 4, x = - 4

Solution steps

(2 x - 3) (3 x - 4) - (x - 13) (x - 4) = 40

Expand

(2 x - 3) (3 x - 4) - (x - 13) (x - 4) : 5 x^{2} - 40

5 x^{2} - 40 = 40

Move

40

to the left side

5 x^{2} - 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 5 ( - 80 )}{2 \cdot 5}

0^{2} - 4 \cdot 5 (- 80) = 40

x_{1, 2} = \frac{- 0 \pm 40}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 0 + 40}{2 \cdot 5}, x_{2} = \frac{- 0 - 40}{2 \cdot 5}

x = \frac{- 0 + 40}{2 \cdot 5} : 4

x = \frac{- 0 - 40}{2 \cdot 5} : - 4

The solutions to the quadratic equation are:

x = 4, x = - 4

2 x^{2} + 2 x = 12

(x - 2)^{2} = 5

- 120 - 2 x = 89 + 9 x

ln (- 10)

What are the solutions to the equation (2x-3)(3x-4)-(x-13)(x-4)=40 ?
The solutions to the equation (2x-3)(3x-4)-(x-13)(x-4)=40 are x=4,x=-4
Find the zeros of (2x-3)(3x-4)-(x-13)(x-4)=40
The zeros of (2x-3)(3x-4)-(x-13)(x-4)=40 are x=4,x=-4