math questionWhat are the solutions of equation 2x^2-8x-192=0?

Asked on by essesy

3 Answers

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation to be solved is : 2x^2-8x-192=0

2x^2-8x-192=0

=> 2x^2 - 24x + 16x - 192 = 0

=> 2x(x - 12) + 16(x - 12) = 0

=> (2x +16)(x - 12) = 0

=> x = -16/2 = -8 and x = 12

The solutions of the equation are x = -8 and x = 12

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

The equation 2x^2-8x-192=0 has to be solved.

2x^2-8x-192=0

Eliminate the common factor 2

x^2 - 4x - 96 = 0

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

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

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

x = -8 and x = 12

The solution of the equation 2x^2-8x-192=0 is x = -8 and x = 12

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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 quadraticÂ in the given equation:

x^2 - 4x - 96 = 0

S = 4 andÂ  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