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

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

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

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

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x, 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

(11 - 2 x) (24 - 2 x) = 114

\frac{( 15 c ^{2} - 8 c )}{3} = 4

x^{2} = x (4 - x)

5 x^{2} - 8 x + 12 = 3 x + 10

(a + 4)^{2} = 2 a (5 a - 1) - 7 (a - 2)

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