Question

Solve the equation 6w2 – 7w – 20 = 0.

Answer 1

For this case we have the following quadratic equation:

$6w ^ 2-7w-20 = 0$

Where:

$a = 6\\b = -7\\c = -20$

Its roots will be given by:

$w = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\w = \frac {- (- 7) \pm \sqrt {(- 7) ^ 2-4 (6) (- 20)}} {2 (6)}\\w = \frac {7 \pm \sqrt {49 + 480}} {12}\\w = \frac {7 \pm \sqrt {529}} {12}\\w = \frac {7 \pm23} {12}$

The roots are:

$w_ {1} = \frac {7 + 23} {12} = \frac {30} {12} = \frac {15} {6} = \frac {5} {2}\\w_ {2} = \frac {7-23} {12} = \frac {-16} {12} = - \frac {8} {6} = - \frac {4} {3}$

Answer:

$w_ {1} = \frac {5} {2}\\w_ {2} = - \frac {4} {3}$

Answer 2

ANSWER

${w} = \frac{5}{2}\: or \: w = - \frac{4}{3}$

EXPLANATION

We want to solve the equation:

$6 {w}^{2} - 7w - 20 = 0$

Let us use the method of factorization

Split the middle term;

$6 {w}^{2} -1 5w + 8w- 20 = 0$

Factor by grouping:

$3w( {w} - 5)w + 4(2w- 5) = 0$

Factor further,

$( {2w} - 5)(3w + 4) = 0$

Use the zero product principle.

$( {2w} - 5) = 0 \: or \: (3w + 4) = 0$

${2w} = 5\: or \: 3w = - 4$

The solution is:

${w} = \frac{5}{2}\: or \: w = - \frac{4}{3}$