You can solve this using the Quadratic Formula, x =[tex] \frac{b (+ or -) \sqrt{ B^{2}-4ac}}{2a} [/tex].
a = 1, b = -12, c = 59
x = [tex] \frac{12 (+ or -) \sqrt{(-12)^{2} -4(1)(59)} }{2(1)} [/tex] Multiply 4 and 1
x = [tex] \frac{12 (+ or -) \sqrt{(-12)^{2} -4(59)} }{2(1)} [/tex] Multiply 2 and 1
x = [tex] \frac{12 (+ or -) \sqrt{(-12)^{2} -4(59)} }{2} [/tex] Multiply -4 and 59
x = [tex] \frac{12 (+ or -) \sqrt{(-12)^{2} -236} }{2} [/tex] Square 12
x = [tex] \frac{12 (+ or -) \sqrt{144 - 236} }{2} [/tex] Subtract
x = [tex] \frac{12 (+ or -) \sqrt{-92} }{2} [/tex] You can't find the square root of negatives, so factor the -92 out.
x = [tex] \frac{12 (+ or -) \sqrt{4 (-23)} }{2} [/tex] You can find the sqaure of 4, so take that out.
x = [tex] \frac{12 (+ or -) 2 \sqrt{-23} }{2} [/tex] Split the expression into two parts
x = [tex] \frac{12}{2} [/tex] (+ or -) [tex] \frac{2 \sqrt{-23} }{2} [/tex] The 2 in the numerator and the 2 in the denominator cancel each other out
x = [tex] \frac{12}{2} [/tex] (+ or -) [tex] \sqrt{-23} [/tex] Divide 12 by 2
x = 6 (+ or -) [tex] \sqrt{-23} [/tex] Now, split the solution into the plus and minus parts.
[tex] \left \{ {{x = 6 + \sqrt{-23} } \atop {x = 6 - \sqrt{-23} }} \right. [/tex]
