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

Question

anonymous · Answer

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:
sqroot(x-5/3) = 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

anonymous · Answer

$$3(x^2-\frac{10}{3}x)=-8$$ $$(\frac{10}{3})^2=\frac{100}{9}......3(x^2-\frac{10}{3}x + \frac{100}{9} )=-8+\frac{300}{9}$$ $$3(x-\frac{10}{3})^2=\frac{76}{3}............\sqrt{(x-\frac{10}{3})}=\frac{\sqrt{76}}{3}$$$$x=\frac{10+/-2\sqrt{19}}{3}$$