What are the solutions to the equation x^2+4x-96=0 ?

The solutions to the equation x^2+4x-96=0 are x=8,x=-12

Find the zeros of x^2+4x-96=0

The zeros of x^2+4x-96=0 are x=8,x=-12

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x^{2} + 4 x - 96 = 0

x = 8, x = - 12

Solution steps

x^{2} + 4 x - 96 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 96) = 20

x_{1, 2} = \frac{- 4 \pm 20}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 20}{2 \cdot 1}, x_{2} = \frac{- 4 - 20}{2 \cdot 1}

x = \frac{- 4 + 20}{2 \cdot 1} : 8

x = \frac{- 4 - 20}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

x = 8, x = - 12

a^{7}

n^{2} + 3 n - 54

15 r^{2} + 7 r - 4 = 0

20 - 4

12 x + 6

What are the solutions to the equation x^2+4x-96=0 ?
The solutions to the equation x^2+4x-96=0 are x=8,x=-12
Find the zeros of x^2+4x-96=0
The zeros of x^2+4x-96=0 are x=8,x=-12