Ok let’s solve it
5(x-2)^2-20=0
first let’s foil (x-2)
5(x^2-4x+4) -20=0
now distribute the 5
5x^2 -20x +20 -20 = 0
combine like terms
5x^2-20x=0
take the gcf
5x(x-4)=0
x=0, 4
solutions are (4,0) and (2, -20) because the original vertex form a(x-h)^2+k
AC OD 4 DE LE 7 ... because you add
4(3h-4)
12h-16?
I don't think that is right, I was kinda guessing
D=LM/(R2+R1)
d(R2+R1)<span>=LM
[</span>d<span>(R2+R1)</span>]/M=L
