What are the solutions to the equation 0=-x^2+4x+12 ?

The solutions to the equation 0=-x^2+4x+12 are x=-2,x=6

Find the zeros of 0=-x^2+4x+12

The zeros of 0=-x^2+4x+12 are x=-2,x=6

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0 = - x^{2} + 4 x + 12

x = - 2, x = 6

Solution steps

0 = - x^{2} + 4 x + 12

Switch sides

- x^{2} + 4 x + 12 = 0

Solve with the quadratic formula

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

4^{2} - 4 (- 1) \cdot 12 = 8

x_{1, 2} = \frac{- 4 \pm 8}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 4 + 8}{2 ( - 1 )}, x_{2} = \frac{- 4 - 8}{2 ( - 1 )}

x = \frac{- 4 + 8}{2 ( - 1 )} : - 2

x = \frac{- 4 - 8}{2 ( - 1 )} : 6

The solutions to the quadratic equation are:

x = - 2, x = 6

3 (x - 2) = 2 (x^{2} - 9) + x^{2} - 5 (x + 2)

- 2 x^{2} + 3 x + 8 = 0

0 = 12 x^{2} - 280 x + 1000

x, x^{2} - 2 x y + y^{2} = 6

(x - 4)^{2} = - 16

What are the solutions to the equation 0=-x^2+4x+12 ?
The solutions to the equation 0=-x^2+4x+12 are x=-2,x=6
Find the zeros of 0=-x^2+4x+12
The zeros of 0=-x^2+4x+12 are x=-2,x=6