Answer:
[tex]\large\boxed{3x^2+6x-5=0\Rightarrow3(x+1)^2-8=0}[/tex]
Step-by-step explanation:
The vertex form of an equation of a quadratic equation
[tex]ax^2+bx+c=0[/tex]:
[tex]a(x-h)^2+k=0[/tex]
(h, k) - vertex
METHOD 1:
Use [tex](a+b)^2=a^2+2ab+b^2\qquad(*)[/tex]
[tex]3x^2+6x-5=0\\\\3x^2+2(x)(3)(1)-5=0\\\\3x^2+(3)(2)(x)(1)+(3)(1^2)-(3)(1^2)-5=0\\\\3\underbrace{(x^2+2(x)(1)+1^2)}_{(*)}-(3)(1)-5=0\\\\3(x+1)^2-8=0[/tex]
METHOD 2:
Use the formula:
[tex]h=\dfrac{-b}{2a},\ k=ah^2+bh+c[/tex]
Substitute:
[tex]3x^2+6x-5\to a=3,\ b=6,\ c=-5\\\\h=\dfrac{-6}{(2)(3)}=\dfrac{-6}{6}=-1\\\\k=3(-1)^2+6(-1)-5=3-6-5=-8[/tex]
[tex]a(x-h)^2+k:\\\\3(x-(-1))^2+(-8)=3(x+1)^2-8[/tex]
