What are the solutions to the equation a^2+14a+48=0 ?

The solutions to the equation a^2+14a+48=0 are a=-6,a=-8

Find the zeros of a^2+14a+48=0

The zeros of a^2+14a+48=0 are a=-6,a=-8

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a^{2} + 14 a + 48 = 0

a = - 6, a = - 8

Solution steps

a^{2} + 14 a + 48 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 1 \cdot 48}{2 \cdot 1}

1 4^{2} - 4 \cdot 1 \cdot 48 = 2

a_{1, 2} = \frac{- 14 \pm 2}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- 14 + 2}{2 \cdot 1}, a_{2} = \frac{- 14 - 2}{2 \cdot 1}

a = \frac{- 14 + 2}{2 \cdot 1} : - 6

a = \frac{- 14 - 2}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

a = - 6, a = - 8

n - 5 \leq 6

\frac{n - 7}{n ^{2} - n - 42}

6 x^{2} - 5 x \leq 2 x + 3

5 (x - 6) - 6 (x - 5) = x + 5 - (x - 6)

5 a^{2} - 125

What are the solutions to the equation a^2+14a+48=0 ?
The solutions to the equation a^2+14a+48=0 are a=-6,a=-8
Find the zeros of a^2+14a+48=0
The zeros of a^2+14a+48=0 are a=-6,a=-8