What are the solutions to the equation (3x-4)^2-6=(2x-5)^2 ?

The solutions to the equation (3x-4)^2-6=(2x-5)^2 are x=(2+sqrt(79))/5 ,x=(2-sqrt(79))/5

Find the zeros of (3x-4)^2-6=(2x-5)^2

The zeros of (3x-4)^2-6=(2x-5)^2 are x=(2+sqrt(79))/5 ,x=(2-sqrt(79))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(3 x - 4)^{2} - 6 = (2 x - 5)^{2}

x = \frac{2 + 79}{5}, x = \frac{2 - 79}{5}

Decimal Notation

x = 2.17763 \dots, x = - 1.37763 \dots

Solution steps

(3 x - 4)^{2} - 6 = (2 x - 5)^{2}

Expand

(3 x - 4)^{2} - 6 : 9 x^{2} - 24 x + 10

Expand

(2 x - 5)^{2} : 4 x^{2} - 20 x + 25

9 x^{2} - 24 x + 10 = 4 x^{2} - 20 x + 25

Move

25

to the left side

9 x^{2} - 24 x - 15 = 4 x^{2} - 20 x

Move

20 x

to the left side

9 x^{2} - 4 x - 15 = 4 x^{2}

Move

4 x^{2}

to the left side

5 x^{2} - 4 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 5 ( - 15 )}{2 \cdot 5}

(- 4)^{2} - 4 \cdot 5 (- 15) = 279

x_{1, 2} = \frac{- ( - 4 ) \pm 2 79}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 79}{2 \cdot 5}, x_{2} = \frac{- ( - 4 ) - 2 79}{2 \cdot 5}

x = \frac{- ( - 4 ) + 2 79}{2 \cdot 5} : \frac{2 + 79}{5}

x = \frac{- ( - 4 ) - 2 79}{2 \cdot 5} : \frac{2 - 79}{5}

The solutions to the quadratic equation are:

x = \frac{2 + 79}{5}, x = \frac{2 - 79}{5}

x, 6 x y - 3 x^{2} = 0

8 x^{2} - 15 x + 10 = 0

15 + 6 x - 9 x^{2}

x^{2} + 9 = 5

(x + 4)^{2} = 3

What are the solutions to the equation (3x-4)^2-6=(2x-5)^2 ?
The solutions to the equation (3x-4)^2-6=(2x-5)^2 are x=(2+sqrt(79))/5 ,x=(2-sqrt(79))/5
Find the zeros of (3x-4)^2-6=(2x-5)^2
The zeros of (3x-4)^2-6=(2x-5)^2 are x=(2+sqrt(79))/5 ,x=(2-sqrt(79))/5