2х2+3х-11=0; х2-14+33=0; х4-10х2+9=0; -3х2+10х-3=0 помогите )

Решите задачу:

$2x^2+3x-11=0 \\ \\ D=b^2-4ac=3^2-4*2*(-11)=9+88=97 \\ \\ x_{1}=\frac{-b+\sqrt{D}}{2a}=\frac{-3+\sqrt{97}}{4} \\ \\ x_{2}=\frac{-b-\sqrt{D}}{2a}=\frac{-3-\sqrt{97}}{4}$

$x^2-14x+33=0 \\ \\ D=b^2-4ac=(-14)^2-4*1*33=196-132=64 \\ \\ x_{1}=\frac{-b+\sqrt{D}}{2a}=\frac{14+8}{2}= \frac{22}{2}=11 \\ \\ x_{2}=\frac{-b-\sqrt{D}}{2a}=\frac{14-8}{2}= \frac{6}{2} =3$

$x^4-10x^2+9=0 \\ x^2=t \\ t^2-10t+9=0 \\ \\ D=b^2-4ac=(-10)^2-4*1*9=100-36=64 \\ \\ t_{1}=\frac{-b+\sqrt{D}}{2a}=\frac{10+8}{2}= \frac{18}{2}=9 \\ \\ t_{2}=\frac{-b-\sqrt{D}}{2a}=\frac{10-8}{2}= \frac{2}{2} =1 \\ \\ x^2=9, x_1=3, x_2=-3 \\ \\ x^2=1, x_3=1, x_4=-1$

$-3x^2+10x-3=0 \\ \\ D=b^2-4ac=10^2-4*(-3)*(-3)=100-36=64 \\ \\ x_{1}=\frac{-b+\sqrt{D}}{2a}=\frac{-10+8}{-6}= \frac{-2}{-6}= \frac{1}{3} \\ \\ x_{2}=\frac{-b-\sqrt{D}}{2a}=\frac{-10-8}{-6}= \frac{-18}{-6} =3$