What are the solutions to the equation x^2+12x-36=0 ?

The solutions to the equation x^2+12x-36=0 are x=6(sqrt(2)-1),x=-6(1+sqrt(2))

Find the zeros of x^2+12x-36=0

The zeros of x^2+12x-36=0 are x=6(sqrt(2)-1),x=-6(1+sqrt(2))

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x^{2} + 12 x - 36 = 0

x = 6 (2 - 1), x = - 6 (1 + 2)

Decimal Notation

x = 2.48528 \dots, x = - 14.48528 \dots

Solution steps

x^{2} + 12 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 36 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 36) = 122

x_{1, 2} = \frac{- 12 \pm 12 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 12 2}{2 \cdot 1}, x_{2} = \frac{- 12 - 12 2}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = 6 (2 - 1), x = - 6 (1 + 2)

\frac{1}{\frac{1}{a} + \frac{1}{b}}

6 s^{3} + 10 s^{2} - 9 s - 15

58 - 18 - 22

3 x - 1 > 8

\frac{5 c - 1}{2} + \frac{3 c - 4}{4} = 18

What are the solutions to the equation x^2+12x-36=0 ?
The solutions to the equation x^2+12x-36=0 are x=6(sqrt(2)-1),x=-6(1+sqrt(2))
Find the zeros of x^2+12x-36=0
The zeros of x^2+12x-36=0 are x=6(sqrt(2)-1),x=-6(1+sqrt(2))