Question

$3x^{2}+7x-2=3* (\frac{-16}{3} )^{2} +7*( \frac{-16}{3}) -2$

подробно, пожалуйста)

Answer 1

$3x^2+7x-2=3*(-\frac{16}{3})^2+7*(-\frac{16}3)-2$
$3x^2+7x-2=3*\frac{256}{9}-\frac{112}3-\frac{6}3$
$3x^2+7x-2=\frac{256-112-6}3$
$3x^2+7x-2=\frac{138}3$
$3x^2+7x-2=46$
$3x^2+7x-48=0$
$D=7*7-4*3*(-48)=625$
$x_1=\frac{-7+25}6=3$
$x_2=\frac{-7-25}6=\frac{-16}3$