How do you solve $5 x^{2} - 10 x - 12 = 0$ $5x^2 - 10x - 12 = 0$ ?

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Answer 1 · 2015-08-08T20:34:58

$x = 1 \pm \frac{\sqrt{85}}{5}$ $x = 1 +- sqrt(85)/5$

$5 x^{2} - 10 x - 12$ $5x^2-10x-12$ is of the form $a x^{2} + b x + c$ $ax^2+bx+c$ with $a = 5$ $a=5$ , $b = - 10$ $b=-10$ and $c = - 12$ $c=-12$

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = (-10)^2-(4xx5xx-12)$

$= 100+240 = 340 = 2^2*85$

This is positive, but not a perfect square, so the quadratic equation has a pair of irrational roots, given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)$

$=(10+-sqrt(340))/10$

$=(10+-2sqrt(85))/10$

$=1+-sqrt(85)/5$

How do you solve 5x^2 - 10x - 12 = 0 5x2−10x−12=05x^2 - 10x - 12 = 0 ?