Решить систему 4x+3y=-1 2x^2-y=11

$\left \{ {{4x+3y=-1} \atop {2x^2-y=11}} \right.$

$\left \{ {{4x+3y=-1} \atop {6x^2-3y=33}} \right.$

$6x^2+4x-32=0$

$3x^2+2x-16=0$

$D=196$

$x_1=-\frac{8}{3}, x_2=2$

$\left \{ {{x=-\frac{8}{3}} \atop {y=2x^2-11}} \right.$ или $\left \{ {{x=2} \atop {y=2x^2-11}} \right.$

$\left \{ {{x=-\frac{8}{3}} \atop {y=\frac{29}{9}}} \right.$ или $\left \{ {{x=2} \atop {y=-3}} \right.$