Answer:
1.) [tex]x_1=\dfrac{7+\sqrt{13}}{2}\,,\qquad x_2=\dfrac{7-\sqrt{13}}{2}[/tex]
2.) [tex]x_1=-\dfrac13\,,\qquad 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-\sqrt{b^2-4ac}}{2a}\\\\x=\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}\,,\qquad 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-\sqrt{b^2-4ac}}{2a}\\\\x=\dfrac{-10\pm\sqrt{10^2-4\cdot3\cdot3}}{2\cdot3}=\dfrac{-10\pm\sqrt{100-36}}{6}=\dfrac{-10\pm\sqrt{64}}{6}\\\\x_1=\dfrac{-10+8}{6}=\dfrac{-2}6=-\dfrac13\,,\qquad x_2=\dfrac{-10-8}{6}=\dfrac{-18}{6}=-3[/tex]
