Question

Porfa ayudenme a resolver esto 12x al cuadrado, +3x+21

Answer 1

Respuesta:

$12x {}^{2} + 3x + 21 = 0 \\ 3(4x {}^{2} + x + 7) = 0 \\ 4x { }^{2} + x + 7 = 0 \\( 4x {}^{2} + x + \frac{1}{16} ) - \frac{1}{16} + 7 = 0 \\ {(2x + \frac{1}{4} )}^{2} = \frac{1}{16} - 7 \\ {(2x + \frac{1}{4} )}^{2} = \frac{ - 111}{16} \\ 2x + \frac{1}{4} = + - \sqrt{ \frac{ - 111}{16} } \\ 2x + \frac{1}{4} = + - ( \frac{ \sqrt{111} }{4}i) \\ x = + - ( \frac{ \sqrt{111} }{8} i) - \frac{1}{8}$