Question

Which are the solutions of the quadratic equation?

Answer 1

For this case we must find the roots of the following equation:

$x ^ 2-7x-4 = 0$

We have to:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Where:

$a = 1\\b = -7\\c = -4$

Substituting the values:

$x = \frac {- (- 7) \pm \sqrt {(- 7) ^ 2-4 (1) (- 4)}} {2 (1)}\\x = \frac {7 \pm \sqrt {49 + 16}} {2}\\x = \frac {7\pm\sqrt {65}} {2}$

We have two roots:

$x_ {1} = \frac {7+ \sqrt {65}} {2} = 7.53\\x_ {2} = \frac {7- \sqrt {65}} {2} = - 0.53$

Answer:

$x_ {1} = \frac {7+ \sqrt {65}} {2} = 7.53\\x_ {2} = \frac {7- \sqrt {65}} {2} = - 0.53$