What are the solutions to the equation -12=2x^2+2x-12 ?

The solutions to the equation -12=2x^2+2x-12 are x=0,x=-1

Find the zeros of -12=2x^2+2x-12

The zeros of -12=2x^2+2x-12 are x=0,x=-1

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

x = 0, x = - 1

Solution steps

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

Switch sides

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

Move

12

to the left side

2 x^{2} + 2 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

2^{2} - 4 \cdot 2 \cdot 0 = 2

x_{1, 2} = \frac{- 2 \pm 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 2}{2 \cdot 2}, x_{2} = \frac{- 2 - 2}{2 \cdot 2}

x = \frac{- 2 + 2}{2 \cdot 2} : 0

x = \frac{- 2 - 2}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

x = 0, x = - 1

x^{2} - x - 4

8 x^{3} + 8 x^{2} - 18 x - 18

8 (2 + 58)

\frac{4 5}{3 2}

10 d^{2} - 29 d + 10 = 0

What are the solutions to the equation -12=2x^2+2x-12 ?
The solutions to the equation -12=2x^2+2x-12 are x=0,x=-1
Find the zeros of -12=2x^2+2x-12
The zeros of -12=2x^2+2x-12 are x=0,x=-1