Question

help

Answer 1

how

Answer 2

exactly

Answer 3

help:) solving by competing the square:
3x^2-10x+8=0

Answer 4

\[-b \pm \sqrt{b^2+2ac}\div2a\]

Answer 5

Divide both sides by 3:
x^2-(10 x)/3+8/3 = 0
Subtract 8/3 from both sides:
x^2-(10 x)/3 = -8/3
Add 25/9 to both sides:
x^2-(10 x)/3+25/9 = 1/9
Factor the left hand side:
(x-5/3)^2 = 1/9
Take the square root of both sides:
sqrt(x-5/3) = (+or -)1/3
Eliminate the absolute value:
x-5/3 = -1/3 or x-5/3 = 1/3
Add 5/3 to both sides:
x = 4/3 or x-5/3 = 1/3
Add 5/3 to both sides:
x = 4/3 or x = 2

Answer 6

sry b^2 - 4ac