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

The solutions to the equation (3x-2)^2=(2x+3)(2x-3)+3(x+1) are x=2,x=1

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

The zeros of (3x-2)^2=(2x+3)(2x-3)+3(x+1) are x=2,x=1

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(3 x - 2)^{2} = (2 x + 3) (2 x - 3) + 3 (x + 1)

x = 2, x = 1

Solution steps

(3 x - 2)^{2} = (2 x + 3) (2 x - 3) + 3 (x + 1)

Expand

(3 x - 2)^{2} : 9 x^{2} - 12 x + 4

Expand

(2 x + 3) (2 x - 3) + 3 (x + 1) : 4 x^{2} + 3 x - 6

9 x^{2} - 12 x + 4 = 4 x^{2} + 3 x - 6

Move

6

to the left side

9 x^{2} - 12 x + 10 = 4 x^{2} + 3 x

Move

3 x

to the left side

9 x^{2} - 15 x + 10 = 4 x^{2}

Move

4 x^{2}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 5 \cdot 10}{2 \cdot 5}

(- 15)^{2} - 4 \cdot 5 \cdot 10 = 5

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

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 5}{2 \cdot 5}, x_{2} = \frac{- ( - 15 ) - 5}{2 \cdot 5}

x = \frac{- ( - 15 ) + 5}{2 \cdot 5} : 2

x = \frac{- ( - 15 ) - 5}{2 \cdot 5} : 1

The solutions to the quadratic equation are:

x = 2, x = 1

S = 2 π^{7} (7 + 11)

p^{- 1} x p^{- 1} x p^{- 1}

- 5 x + 20 < 5

- 12 (x + 5) = 3 (x + 0)

\frac{i ^{5} + 3}{i ^{3} - 1}

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