What are the solutions to the equation -2x^2-2x+40=0 ?

The solutions to the equation -2x^2-2x+40=0 are x=-5,x=4

Find the zeros of -2x^2-2x+40=0

The zeros of -2x^2-2x+40=0 are x=-5,x=4

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- 2 x^{2} - 2 x + 40 = 0

x = - 5, x = 4

Solution steps

- 2 x^{2} - 2 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 2 ) \cdot 40}{2 ( - 2 )}

(- 2)^{2} - 4 (- 2) \cdot 40 = 18

x_{1, 2} = \frac{- ( - 2 ) \pm 18}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 18}{2 ( - 2 )}, x_{2} = \frac{- ( - 2 ) - 18}{2 ( - 2 )}

x = \frac{- ( - 2 ) + 18}{2 ( - 2 )} : - 5

x = \frac{- ( - 2 ) - 18}{2 ( - 2 )} : 4

The solutions to the quadratic equation are:

x = - 5, x = 4

x^{2} = 69

x, x^{2} y^{9} = 2

3 x^{2} + 6 x - 4 = 2 x + 3

x (2 + 2 x) = 60

y^{2} - 4 y + 29 = 0

What are the solutions to the equation -2x^2-2x+40=0 ?
The solutions to the equation -2x^2-2x+40=0 are x=-5,x=4
Find the zeros of -2x^2-2x+40=0
The zeros of -2x^2-2x+40=0 are x=-5,x=4