How do you solve $5x^2+9x-2=0$ using the quadratic formula?

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Answer 1 · 2015-07-25T14:08:03

The solutions for the equation are:
$color(blue)(x=-2,x=1/5$

The equation $5x^2+9x−2$ is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=5, b=9, c=-2$

The Discriminant is given by:
$Delta=b^2-4*a*c$

$= (9)^2 - (4)* (5) * (-2)$

$= 81+40 = 121$

As $Delta>0$ there are two solutions.

The solutions are found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

As $Delta =121$ , $x = ((-9)+-sqrt(121))/(2*5) = (-9+-11)/10$

$x=((-9-11)/10) = -20/10 =color(blue)( -2$

$x=((-9+11)/10) = 2/10 =color(blue)( 1/5$

How do you solve 5x^2+9x-2=05x^2+9x-2=0 using the quadratic formula?