What are the solutions to the equation -x^2-6x+9=0 ?

The solutions to the equation -x^2-6x+9=0 are x=-3(1+sqrt(2)),x=3(sqrt(2)-1)

Find the zeros of -x^2-6x+9=0

The zeros of -x^2-6x+9=0 are x=-3(1+sqrt(2)),x=3(sqrt(2)-1)

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- x^{2} - 6 x + 9 = 0

x = - 3 (1 + 2), x = 3 (2 - 1)

Decimal Notation

x = - 7.24264 \dots, x = 1.24264 \dots

Solution steps

- x^{2} - 6 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 1 ) \cdot 9}{2 ( - 1 )}

(- 6)^{2} - 4 (- 1) \cdot 9 = 62

x_{1, 2} = \frac{- ( - 6 ) \pm 6 2}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 6 2}{2 ( - 1 )}, x_{2} = \frac{- ( - 6 ) - 6 2}{2 ( - 1 )}

x = \frac{- ( - 6 ) + 6 2}{2 ( - 1 )} : - 3 (1 + 2)

x = \frac{- ( - 6 ) - 6 2}{2 ( - 1 )} : 3 (2 - 1)

The solutions to the quadratic equation are:

x = - 3 (1 + 2), x = 3 (2 - 1)

50 x^{2} - 144 = 0

3 x^{2} - 4050 x + 1365000 = 0

t, m t^{2} + n t = z

x, x^{2} + y^{2} + 4 z^{2} = 16

5 t^{2} + 7 t - 3 = 4 t

What are the solutions to the equation -x^2-6x+9=0 ?
The solutions to the equation -x^2-6x+9=0 are x=-3(1+sqrt(2)),x=3(sqrt(2)-1)
Find the zeros of -x^2-6x+9=0
The zeros of -x^2-6x+9=0 are x=-3(1+sqrt(2)),x=3(sqrt(2)-1)