Answer: 3 ± √5
Step-by-step explanation:
[tex]3a-4=a(a-3)\\\\3a-4=a^2-3a\qquad distributed\ "a"\ on\ the\ right\ side\\\\0=a^2-6a+4\qquad subtracted\ 3a-4\ from\ both\ sides\\\\a=1,\ b=-6,\ c=4\\\\\\\text{Quadratic formula is: }x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-6)\pm \sqrt{(-6)^2-4(1)(4)}}{2(1)}\\\\\\.\ =\dfrac{6\pm \sqrt{36-16}}{2}\\\\\\.\ =\dfrac{6\pm \sqrt{20}}{2}\\\\\\,\ =\dfrac{6\pm 2\sqrt5}{2}\\\\\\.\ =3\pm\sqrt5[/tex]
