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

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Answer 1 · 2018-05-07T04:17:16

$x = -3$
$x = -2$

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

for a quadratic of the form $ax^2 + bx +c$ .

We have $a = 1$ , $b = 5$ , and $c = 6$ .

$x=(-5 +- sqrt(5^2-4(1)(6)))/(2(1))$

$x = (-5 +- sqrt(25-24))/(2)$

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

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

Hence,

$x = -6/2 = -3$
and
$x = -4/2 = -2$

I'm not sure why you wanted to use the quadratic formula, but you could just factor the quadratic:

$x^2+5x+6 = (x+3)(x+2) = 0$

$x+3 = 0 -> x = -3$
and
$x+2=0 ->x = -2$

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