# How do you solve x(x+6)=-15 using the quadratic formula?

Oct 25, 2017

No real solutions, x>0

#### Explanation:

Expand boiii

$x \left(x + 6\right)$

$x \cdot x$ = ${x}^{2}$

$x \cdot 6$ = $6 x$

We have -15, bring that over to make + 15

so the equation is ${x}^{2} + 6 x + 15 = 0$

collect terms for the quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 a}$
a is the number in front of ${x}^{2}$
b is the number in front of $x$
c is the number without an $x$, poor lonely number

a = 1
b = 6
c = 15
remember minuses matter, however there are none in this question.

so plug it in
$\frac{- 6 \pm \sqrt{{6}^{2} - 4 \left(1\right) \left(15\right)}}{2 \left(1\right)}$

put into your calculator as is, i'd suggest one from here: https://goo.gl/JtHwiA they are brilliant. (Casio 991ex)

This leaves a negative number in your square root, which means imaginary numbers.. No real solution is obtainable