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

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

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

The zeros of 12x-4-x^2=0 are x=2(3-2sqrt(2)),x=2(3+2sqrt(2))

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

x = 2 (3 - 22), x = 2 (3 + 22)

Decimal

x = 0.34314 \dots, x = 11.65685 \dots

Solution steps

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

1 2^{2} - 4 (- 1) (- 4) = 82

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

Separate the solutions

x_{1} = \frac{- 12 + 8 2}{2 ( - 1 )}, x_{2} = \frac{- 12 - 8 2}{2 ( - 1 )}

x = \frac{- 12 + 8 2}{2 ( - 1 )} : 2 (3 - 22)

x = \frac{- 12 - 8 2}{2 ( - 1 )} : 2 (3 + 22)

The solutions to the quadratic equation are:

x = 2 (3 - 22), x = 2 (3 + 22)

X^{2} = 81

x^{2} + 3 x - 6 = 0

s^{2} + 2 s - 3

s^{2} + 2 s + 6

1459264 x^{2} + 24288.75 x - 2147391.52 = 0

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