How do you solve $d^{2} + 5 d + 6 = 0$ $d^2+5d+6=0$ using the quadratic formula?

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Answer 1 · 2015-08-10T07:43:01

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

$x = - 3$ $color(blue)(x=-3$

$d^{2} + 5 d + 6 = 0$ $d^2+5d+6=0$

The equation is of the form $a d^{2} + b d + c = 0$ $color(blue)(ad^2+bd+c=0$ where:

$a = 1, b = 5, c = 6$ $a=1, b=5, c=6$

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

$= (5)^2-(4*1*6)$

$= 25 - 24 = 1$

As $Delta=0$ there is only one solution.

The solutions are found using the formula:

$x=(-b+-sqrtDelta)/(2*a)$

$x = ((-5)+-sqrt(1))/(2*1) = (-5+-1)/2$

The solutions for the equation are:
$x =(-5+1)/2 , x=-4/2,color(blue)(x=-2$

$x =(-5-1)/2 , x=-6/2,color(blue)(x=-3$

How do you solve d^2+5d+6=0d2+5d+6=0d^2+5d+6=0 using the quadratic formula?