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

The solutions to the equation 0=x^2+x-12 are x=3,x=-4

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

The zeros of 0=x^2+x-12 are x=3,x=-4

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0 = x^{2} + x - 12

x = 3, x = - 4

Solution steps

0 = x^{2} + x - 12

Switch sides

x^{2} + x - 12 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 12) = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 \cdot 1}, x_{2} = \frac{- 1 - 7}{2 \cdot 1}

x = \frac{- 1 + 7}{2 \cdot 1} : 3

x = \frac{- 1 - 7}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 3, x = - 4

7 (x + 2) + 5 \geq - 3 x - 3

\frac{9}{8} m - \frac{1}{6} m = - 4 - \frac{11}{3}

\frac{5}{x - 2} + \frac{3}{2 - x} \geq 1

4 x + 6 = 2 x - 12

x^{2} + 2 x < 0

What are the solutions to the equation 0=x^2+x-12 ?
The solutions to the equation 0=x^2+x-12 are x=3,x=-4
Find the zeros of 0=x^2+x-12
The zeros of 0=x^2+x-12 are x=3,x=-4