Answer:
1) [tex]x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2[/tex]
2) [tex]x_1=-\dfrac13\,,\quad x_2=-3[/tex]
Step-by-step explanation:
1)
[tex]x^2 - 7x + 9 = 0\quad\implies\quad a=1\,,\ b = -7\,,\ c=9\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\cdot1\cdot9}}{2\cdot1}=\dfrac{7\pm\sqrt{49-36}}2\\\\x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2[/tex]
2)
[tex]3x^2 + 10x=-3\\\\3x^2+10x+3=0\quad\implies\quad a=3\,,\ b =10\,,\ c=3\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-10\pm\sqrt{10^2-4\cdot3\cdot3}}{2\cdot3}= \dfrac{-10\pm\sqrt{100-36}}6\\\\x_1=\dfrac{-10+\sqrt{64}}6=\dfrac{-10+8}6=-\dfrac13\,,\qquad x_2=\dfrac{-10-8}6=-3[/tex]
