Question

${x}^{2} + 13x + 6 = 0$
Find the value(s) of x.....[middle term]

Answer 1

${x}^{2} + 13x + 6 = 0 \\ \\ let \: a = 1(coefficient \: of \: {x}^{2}) \\ b = 13(coefficient \: of \: x) \\ c = 6(the \: constant) \\ \\ x = \frac{b - + \sqrt{ {b}^{2} - 4ac } }{2a}$

$x = \frac{13 - + \sqrt{(13)^{2} - (4 \times 1 \times 6)} }{2 \times 1} \\ \\ x = \frac{13 + - \sqrt{169 - 24} }{2} = \frac{13 + - \sqrt{145} }{2}$