Question

Решите квадратные уравнения: а)3x^2-4x-5=0 б)3x^2-4x-1=0

Answer 1

А
$3 x^{2} -4x-5=0$
$D = 4^{2} + 4*5*3=16+60=76$
$x_{1} = \frac{4- \sqrt{76} }{6} = \frac{4-2 \sqrt{19} }{6} = \frac{2-\sqrt{19}}{3}$
$x_{2} = \frac{4+ \sqrt{76} }{6} = \frac{4+2 \sqrt{19} }{6} = \frac{2+\sqrt{19}}{3}$
Б
$3 x^{2} -4x-1=0$
$D = 4 ^{2}+4*3=16+12=28$
$x_{1} = \frac{4+ \sqrt{28} }{6} = \frac{4+2 \sqrt{7} }{6} = \frac{2+\sqrt{7}}{3}$
$x_{2} = \frac{4- \sqrt{28} }{6} = \frac{4-2 \sqrt{7} }{6} = \frac{2-\sqrt{7}}{3}$