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

The solutions to the equation 0=2x^2-2x-5 are x=(1+sqrt(11))/2 ,x=(1-sqrt(11))/2

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

The zeros of 0=2x^2-2x-5 are x=(1+sqrt(11))/2 ,x=(1-sqrt(11))/2

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

x = \frac{1 + 11}{2}, x = \frac{1 - 11}{2}

Decimal

x = 2.15831 \dots, x = - 1.15831 \dots

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 2 (- 5) = 211

x_{1, 2} = \frac{- ( - 2 ) \pm 2 11}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 11}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 11}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 11}{2 \cdot 2} : \frac{1 + 11}{2}

x = \frac{- ( - 2 ) - 2 11}{2 \cdot 2} : \frac{1 - 11}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 11}{2}, x = \frac{1 - 11}{2}

- 4 x^{2} - 13 x - 3 = 0

x^{2} - 2 - 48 = 0

3 x^{2} + 2 x + 2

25 (5 x - 1)^{2} - 40 = - 4

y, x^{2} = 4 y^{2}

What are the solutions to the equation 0=2x^2-2x-5 ?
The solutions to the equation 0=2x^2-2x-5 are x=(1+sqrt(11))/2 ,x=(1-sqrt(11))/2
Find the zeros of 0=2x^2-2x-5
The zeros of 0=2x^2-2x-5 are x=(1+sqrt(11))/2 ,x=(1-sqrt(11))/2