Answer: [tex](x+10)^2+119=0[/tex]
Step-by-step explanation:
For a quadratic equation, the vertex form is given by : [tex]y=a(x-h)^2+k[/tex], where (h, k) is the vertex.
The given quadratic equation: [tex]x^2 + 20x + 28 = 9[/tex]
Subtract 9 from both sides
[tex]=x^2+20x+19=0[/tex]
compare this to [tex]x^2+bx=c[/tex], and add [tex](\frac{b}{2})^2[/tex] both sides
b= 20
[tex]x^2+20+100+19=-100[/tex] [(b/2)²=20/2=10]
[tex]\Rightarrow\ x^2+2(x)(10)+10^2+119=0[/tex]
[tex]\Rightarrow\ (x+10)^2+119=0\ \ \ [\because\ (a+b)^2=a^2+b^2+2ab][/tex]
So, the vertex form : [tex](x+10)^2+119=0[/tex]
