What are the solutions to the equation a^2+11a=-18 ?

The solutions to the equation a^2+11a=-18 are a=-2,a=-9

Find the zeros of a^2+11a=-18

The zeros of a^2+11a=-18 are a=-2,a=-9

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a^{2} + 11 a = - 18

a = - 2, a = - 9

Solution steps

a^{2} + 11 a = - 18

Move

18

to the left side

a^{2} + 11 a + 18 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot 18}{2 \cdot 1}

1 1^{2} - 4 \cdot 1 \cdot 18 = 7

a_{1, 2} = \frac{- 11 \pm 7}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- 11 + 7}{2 \cdot 1}, a_{2} = \frac{- 11 - 7}{2 \cdot 1}

a = \frac{- 11 + 7}{2 \cdot 1} : - 2

a = \frac{- 11 - 7}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

a = - 2, a = - 9

0 \leq y < 4

\frac{9 - 2 d}{4} = 1 - d

\frac{1}{x + 1} < \frac{1}{( x - 2 ) ( x - 3 )}

2 x^{2} + 2 x - 60 = 0

\frac{- ( - 3 ) - 2 ( - 5 )}{( - 5 ) + 4 ( - 3 )}

What are the solutions to the equation a^2+11a=-18 ?
The solutions to the equation a^2+11a=-18 are a=-2,a=-9
Find the zeros of a^2+11a=-18
The zeros of a^2+11a=-18 are a=-2,a=-9