What are the solutions to the equation x^2+20x-96=0 ?

The solutions to the equation x^2+20x-96=0 are x=4,x=-24

Find the zeros of x^2+20x-96=0

The zeros of x^2+20x-96=0 are x=4,x=-24

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x^{2} + 20 x - 96 = 0

x = 4, x = - 24

Solution steps

x^{2} + 20 x - 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot ( - 96 )}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot (- 96) = 28

x_{1, 2} = \frac{- 20 \pm 28}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 20 + 28}{2 \cdot 1}, x_{2} = \frac{- 20 - 28}{2 \cdot 1}

x = \frac{- 20 + 28}{2 \cdot 1} : 4

x = \frac{- 20 - 28}{2 \cdot 1} : - 24

The solutions to the quadratic equation are:

x = 4, x = - 24

9 x^{2} + 12 x + 5 = 10

v, p^{1} v^{1} = p^{2} v^{2}

3 x^{2} - 5 x = 0

x, x^{2} - 3 m x + 8 = 0

(x + \frac{1}{2})^{2} = \frac{16}{9}

What are the solutions to the equation x^2+20x-96=0 ?
The solutions to the equation x^2+20x-96=0 are x=4,x=-24
Find the zeros of x^2+20x-96=0
The zeros of x^2+20x-96=0 are x=4,x=-24