What are the solutions to the equation (2x-3)^2-2(2x-3)-8=0 ?

The solutions to the equation (2x-3)^2-2(2x-3)-8=0 are x= 7/2 ,x= 1/2

Find the zeros of (2x-3)^2-2(2x-3)-8=0

The zeros of (2x-3)^2-2(2x-3)-8=0 are x= 7/2 ,x= 1/2

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(2 x - 3)^{2} - 2 (2 x - 3) - 8 = 0

x = \frac{7}{2}, x = \frac{1}{2}

Decimal Notation

x = 3.5, x = 0.5

Solution steps

(2 x - 3)^{2} - 2 (2 x - 3) - 8 = 0

Expand

(2 x - 3)^{2} - 2 (2 x - 3) - 8 : 4 x^{2} - 16 x + 7

4 x^{2} - 16 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 4 \cdot 7}{2 \cdot 4}

(- 16)^{2} - 4 \cdot 4 \cdot 7 = 12

x_{1, 2} = \frac{- ( - 16 ) \pm 12}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 12}{2 \cdot 4}, x_{2} = \frac{- ( - 16 ) - 12}{2 \cdot 4}

x = \frac{- ( - 16 ) + 12}{2 \cdot 4} : \frac{7}{2}

x = \frac{- ( - 16 ) - 12}{2 \cdot 4} : \frac{1}{2}

The solutions to the quadratic equation are:

x = \frac{7}{2}, x = \frac{1}{2}

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

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

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

5 x^{2} + 7 x - 5 = 0

5 x^{2} - 8 x + 7 = 0

What are the solutions to the equation (2x-3)^2-2(2x-3)-8=0 ?
The solutions to the equation (2x-3)^2-2(2x-3)-8=0 are x= 7/2 ,x= 1/2
Find the zeros of (2x-3)^2-2(2x-3)-8=0
The zeros of (2x-3)^2-2(2x-3)-8=0 are x= 7/2 ,x= 1/2