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

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

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

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

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x, 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

x, y (x^{2} - 4) = 1 - x^{2}

6 x^{2} - 135 = 3 x^{2}

3 x^{2} + 36 x + 1

5 x^{2} - 10 x + 19 = 0

y, x^{3} + 3 x y^{2} = 1

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