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

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

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

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

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

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

Decimal Notation

d = 2.87082 \dots, d = - 0.87082 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

3 q^{2} + q - 11 = - 4

(x + 24) \times (x + 2) = 420

0 = x^{2} + 5 x - 24

- 12 x - 9 = 4 x^{2}

x, z = 1 - x^{2} - y^{2}

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