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

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

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

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

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x^{2} + 12 x = - 36

x = - 6

Solution steps

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

f (x) = \frac{4 x + 7}{x ^{2} + 12 x + 27}

f (x) = 10 x + 7

f (x) = x + 1

h (x) = 20 x^{2} + 7 x - 3

y = lo g_{3} (x) + 6

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