quadratic formula
for ax^2+bx+c=0
x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
first conver to ax^2+bx+c=0 form
3x^2+5x=8
minus 8 both sides
3x^2+5x-8=0
we can actually solve by factoring (x-1)(3x+8), but that's isn't the question
so
ax^2+bx+c=0
3x^2+5x-8=0
a=3
b=5
c=-8
x=[tex] \frac{-5+/- \sqrt{5^2-4(3)(-8)} }{2(3)} [/tex]
x=[tex] \frac{-5+/- \sqrt{25+96} }{6} [/tex]
x=[tex] \frac{-5+/- \sqrt{121} }{6} [/tex]
x=[tex] \frac{-5+/-11}{6} [/tex]
so
x=(-5-11)/6 or (-5+11)/6
simplify
(-5-11)/6=-16/6=-8/3
(-5+11)/6=6/6=1
x=-8/3 and 1
