What are the solutions to the equation x^2+8x-48=0 ?

The solutions to the equation x^2+8x-48=0 are x=4,x=-12

Find the zeros of x^2+8x-48=0

The zeros of x^2+8x-48=0 are x=4,x=-12

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x^{2} + 8 x - 48 = 0

x = 4, x = - 12

Solution steps

x^{2} + 8 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 48) = 16

x_{1, 2} = \frac{- 8 \pm 16}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 16}{2 \cdot 1}, x_{2} = \frac{- 8 - 16}{2 \cdot 1}

x = \frac{- 8 + 16}{2 \cdot 1} : 4

x = \frac{- 8 - 16}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

x = 4, x = - 12

x^{\frac{1}{3}} \cdot x^{\frac{1}{7}}

18 x^{2} y - 24 x^{2} - 27 y + 36

24 3^{\frac{1}{5}}

\frac{2 x ^{2} - 7 x + 8}{x ^{2} + 2} > 1

81 x^{2} - 72 x + 16

What are the solutions to the equation x^2+8x-48=0 ?
The solutions to the equation x^2+8x-48=0 are x=4,x=-12
Find the zeros of x^2+8x-48=0
The zeros of x^2+8x-48=0 are x=4,x=-12