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