How do you solve using the quadratic formula $5x^2-8x-14=0$ ?

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Answer 1 · 2015-05-10T18:55:28

$5x^2-8x-14=0$ is of the form $ax^2+bx+c=0$ , with $a=5$ , $b = -8$ and $c = -14$ .

The quadratic formula gives $x = (-b +-sqrt(b^2-4ac))/(2a)$ .

So substituting our values for $a$ , $b$ and $c$ , we get:

$x = (8+-sqrt(8^2-4*5*(-14)))/(2*5)$

$= (8+-sqrt(4*16+4*70))/10$

$=(8+-sqrt(4*86))/10$

$=(8+-sqrt(4)*sqrt(86))/10$

$=(8+-2*sqrt(86))/10$

$=(4+-sqrt(86))/5$

$=4/5+-sqrt(86)/5$

How do you solve using the quadratic formula 5x^2-8x-14=05x^2-8x-14=0?