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

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

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

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

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

x = \frac{5}{2}, x = \frac{3}{2}

Solution steps

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

Expand

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

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

Move

6

to the left side

4 x^{2} - 12 x + 15 = 4 x

Move

4 x

to the left side

4 x^{2} - 16 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 4 \cdot 15}{2 \cdot 4}

(- 16)^{2} - 4 \cdot 4 \cdot 15 = 4

x_{1, 2} = \frac{- ( - 16 ) \pm 4}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 4}{2 \cdot 4}, x_{2} = \frac{- ( - 16 ) - 4}{2 \cdot 4}

x = \frac{- ( - 16 ) + 4}{2 \cdot 4} : \frac{5}{2}

x = \frac{- ( - 16 ) - 4}{2 \cdot 4} : \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{5}{2}, x = \frac{3}{2}

- \frac{5}{6} \cdot \frac{1}{6}

- 9 \leq - 1 - t

\frac{3}{4} x + \frac{1}{5} \leq \frac{1}{20}

- 5 x < 35

x^{3} + 2 x^{2} - 6 x - 2 = x^{3} + x^{2} - 7 x + 18

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