What are the solutions to the equation 2x^2-4x-5=0 ?

The solutions to the equation 2x^2-4x-5=0 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2

Find the zeros of 2x^2-4x-5=0

The zeros of 2x^2-4x-5=0 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2

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

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 = 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}

10 a^{4} - a^{3} - 21 a^{2}

(x - 2)^{3} - x (x - 5) \cdot (x - 1) = 2 - 3 x

(1 + i)^{8}

n^{3} + 5 n^{2} - 9 n - 45

1 + 6 x + 9 = 2 (3 x + 3) - x

What are the solutions to the equation 2x^2-4x-5=0 ?
The solutions to the equation 2x^2-4x-5=0 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2
Find the zeros of 2x^2-4x-5=0
The zeros of 2x^2-4x-5=0 are x=(2+sqrt(14))/2 ,x=(2-sqrt(14))/2