Odpowiedź :
a)
[tex]x^2+8x=0\\x(x+8)=0\\\begin{array}{c}x=0\end{array}\lor\begin{array}{c}x+8=0\\x=-8\end{array}[/tex]
b)
[tex]3x^2-\cfrac{9}{4}x=0\\3x\left(x-\cfrac{3}{4}\right)=0\\\begin{array}{c}3x=0\\x=0\end{array}\lor\begin{array}{c}x-\cfrac{3}{4}=0\\x=\cfrac{3}{4}\end{array}[/tex]
c)
[tex]\cfrac{3}{4}y^2+9=\cfrac{y}{2}+(y-3)^2\\\cfrac{3}{4}y^2+9=\cfrac{y}{2}+y^2-6y+9\\\cfrac{1}{4}y^2-\cfrac{11}{2}y=0\\\cfrac{1}{2}y\left(\cfrac{1}{2}y-11\right)=0\\\begin{array}{c}\cfrac{1}{2}y=0\\y=0\end{array}\lor\begin{array}{c}\cfrac{1}{2}y-11=0\\\cfrac{1}{2}y=11\\y=22\end{array}[/tex]
d)
[tex]-2x^2-8=0\\-2x^2=8\\x^2=-4\\x\not\in\mathbb{R}[/tex]
e)
[tex]x^2-2>0\\x^2>2\\x>\sqrt{2}\land x<-\sqrt{2}\\x\in\left(-\infty;-\sqrt{2}\right)\cup\left(\sqrt{2};\infty\right)[/tex]
f)
[tex]x^2+12x=2(x-3)^2\\x^2+12x=2(x^2-6x+9)\\x^2+12x=2x^2-12x+18\\x^2-24x+18=0\\\Delta=b^2-4ac=\left(-24\right)^2-4*1*18=576-72=504\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{24-\sqrt{504}}{2*1}=\cfrac{24-6\sqrt{14}}{2}=12-3\sqrt{14}\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{24+\sqrt{504}}{2*1}=\cfrac{24+6\sqrt{14}}{2}=12+3\sqrt{14}[/tex]
g)
[tex]x^2+169=0\\x^2=-169\\x\not\in\mathbb{R}[/tex]
h)
[tex]5x^2=99-6x^2\\11x^2=99\\x^2=9\\x=-3\lor x=3[/tex]