What are the solutions to the equation x^2+6x=18 ?

The solutions to the equation x^2+6x=18 are x=3(sqrt(3)-1),x=-3(1+sqrt(3))

Find the zeros of x^2+6x=18

The zeros of x^2+6x=18 are x=3(sqrt(3)-1),x=-3(1+sqrt(3))

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x^{2} + 6 x = 18

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

Decimal Notation

x = 2.19615 \dots, x = - 8.19615 \dots

Solution steps

x^{2} + 6 x = 18

Move

18

to the left side

x^{2} + 6 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot (- 18) = 63

x_{1, 2} = \frac{- 6 \pm 6 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 6 + 6 3}{2 \cdot 1}, x_{2} = \frac{- 6 - 6 3}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

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

5 h - 4 \geq 6 and 7 h + 11 < 32

∣ 3 x - 1 ∣ = ∣ x + 2 ∣

7 (x - 3) = 5 x - 1

- y^{4} + 16

2^{- n} (2^{n} - 2^{1 + n})

What are the solutions to the equation x^2+6x=18 ?
The solutions to the equation x^2+6x=18 are x=3(sqrt(3)-1),x=-3(1+sqrt(3))
Find the zeros of x^2+6x=18
The zeros of x^2+6x=18 are x=3(sqrt(3)-1),x=-3(1+sqrt(3))