Question

Solve x^2-12x+59=0 for x

Answer 1

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.

$\left \{ {{x = 6 + \sqrt{-23} } \atop {x = 6 - \sqrt{-23} }} \right.$

Answer 2

Answer:

\left \{ {{x = 6 +  \sqrt{-23} } \atop {x = 6 -  \sqrt{-23} }} \right.

Step-by-step explanation: