What are the solutions to the equation 2x^2-4x=5 ?

The solutions to the equation 2x^2-4x=5 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2

Find the zeros of 2x^2-4x=5

The zeros of 2x^2-4x=5 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2

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2 x^{2} - 4 x = 5

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

Decimal Notation

x = 2.87082 \dots, x = - 0.87082 \dots

Solution steps

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}

ab x^{2} + (3 a + 2 b) x + 6 = 0

x, x^{2} + 10 x + y^{2} + 18 y + 110 = 0

2 x^{2} + 3 x - 18 = 0

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

v, \frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}

What are the solutions to the equation 2x^2-4x=5 ?
The solutions to the equation 2x^2-4x=5 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2
Find the zeros of 2x^2-4x=5
The zeros of 2x^2-4x=5 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2