Question

Plz help me!!!!!!!!!

Answer 1

Answer:   3 ± √5

Step-by-step explanation:

$3a-4=a(a-3)\\\\3a-4=a^2-3a\qquad distributed\ "a"\ on\ the\ right\ side\\\\0=a^2-6a+4\qquad subtracted\ 3a-4\ from\ both\ sides\\\\a=1,\ b=-6,\ c=4\\\\\\\text{Quadratic formula is: }x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-6)\pm \sqrt{(-6)^2-4(1)(4)}}{2(1)}\\\\\\.\ =\dfrac{6\pm \sqrt{36-16}}{2}\\\\\\.\ =\dfrac{6\pm \sqrt{20}}{2}\\\\\\,\ =\dfrac{6\pm 2\sqrt5}{2}\\\\\\.\ =3\pm\sqrt5$

Answer 2

Answer:

Step-by-step explanation:

3a-4=a(a-3)

3a-4=a²-3a

3a=a²-3a+4

0=a²-6a+4

a=(6±√[36-4(1)(4)])/2(1)

a=(6±√(36-16))/2

=(6±√20)/2

=3±(2√5)/2

=3±√5

a=3+√5

a=3-√5