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

The solutions to the equation 2(2x-3)^2-2(2x-3)-2=0 are x=(7+sqrt(5))/4 ,x=(7-sqrt(5))/4

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

The zeros of 2(2x-3)^2-2(2x-3)-2=0 are x=(7+sqrt(5))/4 ,x=(7-sqrt(5))/4

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

x = \frac{7 + 5}{4}, x = \frac{7 - 5}{4}

Decimal

x = 2.30901 \dots, x = 1.19098 \dots

Solution steps

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

Expand

2 (2 x - 3)^{2} - 2 (2 x - 3) - 2 : 8 x^{2} - 28 x + 22

8 x^{2} - 28 x + 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 8 \cdot 22}{2 \cdot 8}

(- 28)^{2} - 4 \cdot 8 \cdot 22 = 45

x_{1, 2} = \frac{- ( - 28 ) \pm 4 5}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 28 ) + 4 5}{2 \cdot 8}, x_{2} = \frac{- ( - 28 ) - 4 5}{2 \cdot 8}

x = \frac{- ( - 28 ) + 4 5}{2 \cdot 8} : \frac{7 + 5}{4}

x = \frac{- ( - 28 ) - 4 5}{2 \cdot 8} : \frac{7 - 5}{4}

The solutions to the quadratic equation are:

x = \frac{7 + 5}{4}, x = \frac{7 - 5}{4}

2 x^{(2)} + 3 x = 0

x^{2} + 282 8^{2} = 535 3^{2}

2 x^{2} - (4 x^{2} - 4 x - 4) = 1160

(x + 2)^{2} + x^{2} = 5 8^{2}

- 10 (- 3 x^{2} - 1) = 0

What are the solutions to the equation 2(2x-3)^2-2(2x-3)-2=0 ?
The solutions to the equation 2(2x-3)^2-2(2x-3)-2=0 are x=(7+sqrt(5))/4 ,x=(7-sqrt(5))/4
Find the zeros of 2(2x-3)^2-2(2x-3)-2=0
The zeros of 2(2x-3)^2-2(2x-3)-2=0 are x=(7+sqrt(5))/4 ,x=(7-sqrt(5))/4