What are the solutions to the equation 2-(2x^2+2x)=-10 ?

The solutions to the equation 2-(2x^2+2x)=-10 are x=-3,x=2

Find the zeros of 2-(2x^2+2x)=-10

The zeros of 2-(2x^2+2x)=-10 are x=-3,x=2

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2 - (2 x^{2} + 2 x) = - 10

x = - 3, x = 2

Solution steps

2 - (2 x^{2} + 2 x) = - 10

Expand

2 - (2 x^{2} + 2 x) : 2 - 2 x^{2} - 2 x

2 - 2 x^{2} - 2 x = - 10

Move

10

to the left side

- 2 x^{2} - 2 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 2 ) \cdot 12}{2 ( - 2 )}

(- 2)^{2} - 4 (- 2) \cdot 12 = 10

x_{1, 2} = \frac{- ( - 2 ) \pm 10}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 10}{2 ( - 2 )}, x_{2} = \frac{- ( - 2 ) - 10}{2 ( - 2 )}

x = \frac{- ( - 2 ) + 10}{2 ( - 2 )} : - 3

x = \frac{- ( - 2 ) - 10}{2 ( - 2 )} : 2

The solutions to the quadratic equation are:

x = - 3, x = 2

- 2 x^{2} + 7 x - 4 = 0

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

x^{2} + \frac{3}{3} x + \frac{7}{3} = 0

- 2 x^{2} - 17 x - 53 = 0

x, 3 y^{2} - 12 z^{2} - 4 x^{2} = 12

What are the solutions to the equation 2-(2x^2+2x)=-10 ?
The solutions to the equation 2-(2x^2+2x)=-10 are x=-3,x=2
Find the zeros of 2-(2x^2+2x)=-10
The zeros of 2-(2x^2+2x)=-10 are x=-3,x=2