What are the solutions to the equation -x^2+12x-5=0 ?

The solutions to the equation -x^2+12x-5=0 are x=6-sqrt(31),x=6+sqrt(31)

Find the zeros of -x^2+12x-5=0

The zeros of -x^2+12x-5=0 are x=6-sqrt(31),x=6+sqrt(31)

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- x^{2} + 12 x - 5 = 0

x = 6 - 31, x = 6 + 31

Decimal Notation

x = 0.43223 \dots, x = 11.56776 \dots

Solution steps

- x^{2} + 12 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 1 ) ( - 5 )}{2 ( - 1 )}

1 2^{2} - 4 (- 1) (- 5) = 231

x_{1, 2} = \frac{- 12 \pm 2 31}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 12 + 2 31}{2 ( - 1 )}, x_{2} = \frac{- 12 - 2 31}{2 ( - 1 )}

x = \frac{- 12 + 2 31}{2 ( - 1 )} : 6 - 31

x = \frac{- 12 - 2 31}{2 ( - 1 )} : 6 + 31

The solutions to the quadratic equation are:

x = 6 - 31, x = 6 + 31

(3 x - 4)^{2} - 1 = 24

x, x^{2} - 2.5 x + 1.5 = 0

8 x^{2} + 32 x - 10 = 0

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

- x^{2} + 8 x + 9 = 0

What are the solutions to the equation -x^2+12x-5=0 ?
The solutions to the equation -x^2+12x-5=0 are x=6-sqrt(31),x=6+sqrt(31)
Find the zeros of -x^2+12x-5=0
The zeros of -x^2+12x-5=0 are x=6-sqrt(31),x=6+sqrt(31)