Question

1. Solve using quadratic formula a) x^2 + 12x - 13 = 0

Answer 1

Here we go!

The quadratic formula: x = [-b ± √(b2 - 4ac)]/2a
a = 1 (understood coefficient)
b = 12
c = -13
$x= \frac{-b\pm \sqrt{b^2-4ac} }{2a} \\ x= \frac{-12\pm \sqrt{12^2-4(1)(-13)} }{2(1)} \\ x= \frac{-12\pm \sqrt{144-(-52)} }{2} \\ x=\frac{-12\pm \sqrt{196} }{2} \\ x=\frac{-12\pm14 }{2}$

Okay, so here, we split into 2 problems. If 14 were positive:
$x=\frac{-12+14 }{2} \\ x= \frac{2}{2} \\ x=1$
If 14 were negative:
$x=\frac{-12-14 }{2} \\ \\ x= \frac {-26}{2} \\ x=-14$

So, $\left \{ {{x=1} \atop {x=-14}} \right.$