What are the solutions to the equation -2x^2-4x+6=0 ?

The solutions to the equation -2x^2-4x+6=0 are x=-3,x=1

Find the zeros of -2x^2-4x+6=0

The zeros of -2x^2-4x+6=0 are x=-3,x=1

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- 2 x^{2} - 4 x + 6 = 0

x = - 3, x = 1

Solution steps

- 2 x^{2} - 4 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 2 ) \cdot 6}{2 ( - 2 )}

(- 4)^{2} - 4 (- 2) \cdot 6 = 8

x_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8}{2 ( - 2 )}, x_{2} = \frac{- ( - 4 ) - 8}{2 ( - 2 )}

x = \frac{- ( - 4 ) + 8}{2 ( - 2 )} : - 3

x = \frac{- ( - 4 ) - 8}{2 ( - 2 )} : 1

The solutions to the quadratic equation are:

x = - 3, x = 1

2^{x} = 7

\frac{3 3}{3}

7 x + 3 \geq (x^{2} + 2)

\frac{1}{3} + \frac{2}{3} - \frac{7}{3}

\frac{( \frac{3 x + y}{2 x y} )}{( \frac{5 x}{x + 2} )}

What are the solutions to the equation -2x^2-4x+6=0 ?
The solutions to the equation -2x^2-4x+6=0 are x=-3,x=1
Find the zeros of -2x^2-4x+6=0
The zeros of -2x^2-4x+6=0 are x=-3,x=1