What are the solutions to the equation -x^2+8x-12=0 ?

The solutions to the equation -x^2+8x-12=0 are x=2,x=6

Find the zeros of -x^2+8x-12=0

The zeros of -x^2+8x-12=0 are x=2,x=6

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- x^{2} + 8 x - 12 = 0

x = 2, x = 6

Solution steps

- x^{2} + 8 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 1 ) ( - 12 )}{2 ( - 1 )}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2, x = 6

w^{2} - 31 w = 132

(6 - i) + (7 + 3 i)

16 g^{2} - g^{3}

56 b + 14 c

∣ x + 7 ∣ < 5

What are the solutions to the equation -x^2+8x-12=0 ?
The solutions to the equation -x^2+8x-12=0 are x=2,x=6
Find the zeros of -x^2+8x-12=0
The zeros of -x^2+8x-12=0 are x=2,x=6