What are the solutions to the equation -1x^2+1x+12=0 ?

The solutions to the equation -1x^2+1x+12=0 are x=-3,x=4

Find the zeros of -1x^2+1x+12=0

The zeros of -1x^2+1x+12=0 are x=-3,x=4

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- 1 x^{2} + 1 x + 12 = 0

x = - 3, x = 4

Solution steps

- 1 \cdot x^{2} + 1 \cdot x + 12 = 0

Refine

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 3, x = 4

(x + 3) (x - 0) = 0

4 = b^{2}

x^{2} + 24 x - 140 = 0

6 x^{2} - 23 x = - 7

(2 x - 3)^{2} = 4 x + 5

What are the solutions to the equation -1x^2+1x+12=0 ?
The solutions to the equation -1x^2+1x+12=0 are x=-3,x=4
Find the zeros of -1x^2+1x+12=0
The zeros of -1x^2+1x+12=0 are x=-3,x=4