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

The solutions to the equation 2(x^2-2x)=5 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2

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

The zeros of 2(x^2-2x)=5 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2

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2 (x^{2} - 2 x) = 5

x = \frac{2 + 14}{2}, x = \frac{2 - 14}{2}

Solution steps

2 (x^{2} - 2 x) = 5

Expand

2 (x^{2} - 2 x) : 2 x^{2} - 4 x

2 x^{2} - 4 x = 5

Move

5

to the left side

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 (- 5) = 214

x_{1, 2} = \frac{- ( - 4 ) \pm 2 14}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 14}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 2 14}{2 \cdot 2}

x = \frac{- ( - 4 ) + 2 14}{2 \cdot 2} : \frac{2 + 14}{2}

x = \frac{- ( - 4 ) - 2 14}{2 \cdot 2} : \frac{2 - 14}{2}

The solutions to the quadratic equation are:

x = \frac{2 + 14}{2}, x = \frac{2 - 14}{2}

x^{2} - 13 + 40 = 0

- x^{2} + 2 x = 3

0.0057 v^{2} + 0.13 v - 100 = 0

v^{2} + 10 v + 24 = 0

(2.629 - x) (0.458 - x) - 0.80 1^{2} = 0

What are the solutions to the equation 2(x^2-2x)=5 ?
The solutions to the equation 2(x^2-2x)=5 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2
Find the zeros of 2(x^2-2x)=5
The zeros of 2(x^2-2x)=5 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2