What are the solutions to the equation x^2+16x+24=0 ?

The solutions to the equation x^2+16x+24=0 are x=2(sqrt(10)-4),x=-2(4+sqrt(10))

Find the zeros of x^2+16x+24=0

The zeros of x^2+16x+24=0 are x=2(sqrt(10)-4),x=-2(4+sqrt(10))

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x^{2} + 16 x + 24 = 0

x = 2 (10 - 4), x = - 2 (4 + 10)

Solution steps

x^{2} + 16 x + 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

1 6^{2} - 4 \cdot 1 \cdot 24 = 410

x_{1, 2} = \frac{- 16 \pm 4 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 16 + 4 10}{2 \cdot 1}, x_{2} = \frac{- 16 - 4 10}{2 \cdot 1}

x = \frac{- 16 + 4 10}{2 \cdot 1} : 2 (10 - 4)

x = \frac{- 16 - 4 10}{2 \cdot 1} : - 2 (4 + 10)

The solutions to the quadratic equation are:

x = 2 (10 - 4), x = - 2 (4 + 10)

∣ x ∣ = 2

\frac{9}{100}

(- 3 x)^{- 1} (6 + 2 x)

\frac{x - 2}{x - 7} > 0

(4 a^{2})^{3}

What are the solutions to the equation x^2+16x+24=0 ?
The solutions to the equation x^2+16x+24=0 are x=2(sqrt(10)-4),x=-2(4+sqrt(10))
Find the zeros of x^2+16x+24=0
The zeros of x^2+16x+24=0 are x=2(sqrt(10)-4),x=-2(4+sqrt(10))