How do you solve $2x^2+3x-5=0$ using the quadratic formula?

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Answer 1 · 2015-08-10T07:19:25

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

$2x^2+3x−5=0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=2, b=3, c=-5$

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

$= (3)^2-(4*2* (-5))$

$= 9+40$

$=49$

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

$x = ((-3)+-sqrt(49))/(2*2) = ((-3+-7))/4$

$x = (-3-7)/4, color(blue)(x=-5/2$

$x=(-3+7)/4, color(blue)(x=1$

How do you solve 2x^2+3x-5=02x^2+3x-5=0 using the quadratic formula?