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

The solutions to the equation -36=x^2+12x are x=-6

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

The zeros of -36=x^2+12x are x=-6

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- 36 = x^{2} + 12 x

x = - 6

Solution steps

- 36 = x^{2} + 12 x

Switch sides

x^{2} + 12 x = - 36

Move

36

to the left side

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 = 0

x_{1, 2} = \frac{- 12 \pm 0}{2 \cdot 1}

x = \frac{- 12}{2 \cdot 1}

\frac{- 12}{2 \cdot 1} = - 6

x = - 6

The solution to the quadratic equation is:

x = - 6

j^{2} + 8 j + 7

2 x - 5 = x + 22

(\frac{8}{27})^{\frac{1}{3}}

30 x^{2} + 13 x - 10

\frac{5}{6} (x + 9) + \frac{3}{4} (x + 1) - \frac{7}{9} = 8

What are the solutions to the equation -36=x^2+12x ?
The solutions to the equation -36=x^2+12x are x=-6
Find the zeros of -36=x^2+12x
The zeros of -36=x^2+12x are x=-6