# What are the solutions of equation 2x^2-8x-192=0?

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First, simplify the equation by doing common factor. So the equation becomes 2(x^2 - 4x - 96) = 0.

To get the solutions of the equation, we just have to factor (x^2 - 4x - 96) = 0,

(x - 12)(x + 8) = 0

Equating the two factors to 0,

x - 12 = 0, x = 12

x + 8 = 0, x = -8

Be sure to check the solutions by substituting -8 & 12 to the original equation.

What are the solutions of equation 2x^2-8x-192=0.

We first divide the equation by 2 keep it simple.

x^2-4x-96 = 0. To solve this we factorise the left and equate each factor to zero, since the right is zero.

x^2-4x-96 = 0.

=> x^2 -12x +8x-96 = 0.

=> x(x-12)+8(x-12) = 0.

=> (x-12)(x+8) = 0.

=> x-12 = 0 or x+8 = 0.

=> x=12 or x= -8.

So the solutions to x are 12 or -8.

First, we'll factorize by 2 the equation:

2(x^2 - 4x - 96) = 0

We'll divide by 2:

x^2 - 4x - 96 = 0

We'll identify the sum and product of the roots of the quadratic in the given equation:

x^2 - 4x - 96 = 0

S = 4 and P = -96

We'll apply the quadratic formula:

x1 = [4 + sqrt(16 + 384)]/2

x1 = (4+20)/2

x1 = 12

x2 = (4-20)/2

x2 = -8

**The solutions of the given quadratic are: x1 = 12 and x2 = -8**