What are the solutions to the equation 1x^2+6x=9 ?

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

Find the zeros of 1x^2+6x=9

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

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1 x^{2} + 6 x = 9

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

Decimal Notation

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

Solution steps

1 \cdot x^{2} + 6 x = 9

Refine

x^{2} + 6 x = 9

Move

9

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

y, 18 y^{2} = (x - 4) \cdot (x - 10)^{2}

7 x^{2} - 12 x + 18 = 0

a^{2} \cdot (x - b) - b^{2} \cdot (x - a) = x^{2} (a - b)

x, 6 x^{2} - x y - 12 y^{2} - 8 x + 29 y - 14 = 0

x^{2} + 3 = 311

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