What are the solutions to the equation 0=2x^2+2x-12 ?

The solutions to the equation 0=2x^2+2x-12 are x=2,x=-3

Find the zeros of 0=2x^2+2x-12

The zeros of 0=2x^2+2x-12 are x=2,x=-3

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

x = 2, x = - 3

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- 2 + 10}{2 \cdot 2} : 2

x = \frac{- 2 - 10}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

x = 2, x = - 3

15 x^{2} - 18 x + 4 = 0

9 b^{2} + 0.27 b = 0

\frac{m}{7} = \frac{5 m}{35}

5 - (3 x + 1) = 6

- \frac{1}{( 1 + 2 ) ^{2}}

What are the solutions to the equation 0=2x^2+2x-12 ?
The solutions to the equation 0=2x^2+2x-12 are x=2,x=-3
Find the zeros of 0=2x^2+2x-12
The zeros of 0=2x^2+2x-12 are x=2,x=-3