What are the solutions to the equation 0=-2x^2-2x-3 ?

The solutions to the equation 0=-2x^2-2x-3 are x=-1/2-i(sqrt(5))/2 ,x=-1/2+i(sqrt(5))/2

Find the zeros of 0=-2x^2-2x-3

The zeros of 0=-2x^2-2x-3 are x=-1/2-i(sqrt(5))/2 ,x=-1/2+i(sqrt(5))/2

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

x = - \frac{1}{2} - i \frac{5}{2}, x = - \frac{1}{2} + i \frac{5}{2}

Solution steps

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

Switch sides

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 2 ) ( - 3 )}{2 ( - 2 )}

Simplify

(- 2)^{2} - 4 (- 2) (- 3) : 25 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 5 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 5 i}{2 ( - 2 )}, x_{2} = \frac{- ( - 2 ) - 2 5 i}{2 ( - 2 )}

x = \frac{- ( - 2 ) + 2 5 i}{2 ( - 2 )} : - \frac{1}{2} - i \frac{5}{2}

x = \frac{- ( - 2 ) - 2 5 i}{2 ( - 2 )} : - \frac{1}{2} + i \frac{5}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} - i \frac{5}{2}, x = - \frac{1}{2} + i \frac{5}{2}

3 x^{2} + 9 x + 12 = 0

x, x^{2} - x yz = (x - yz)

5 y^{2} - 2 = - 9 y

4 y^{2} + 9 = - 12 y

2 - y^{2} = 0

What are the solutions to the equation 0=-2x^2-2x-3 ?
The solutions to the equation 0=-2x^2-2x-3 are x=-1/2-i(sqrt(5))/2 ,x=-1/2+i(sqrt(5))/2
Find the zeros of 0=-2x^2-2x-3
The zeros of 0=-2x^2-2x-3 are x=-1/2-i(sqrt(5))/2 ,x=-1/2+i(sqrt(5))/2