Question

3x^2 + 19x = 154 quadratic formula

Answer 1

for ax²+bx+c=0
x=$\frac{-b+/- \sqrt{b^2-4ac} }{2a}$
given

3x²+19x=154
minus 154 both sides
3x²+19x-154=0
a=3
b=19
c=-154

subsiture

x=$\frac{-(19)+/- \sqrt{(19)^2-4(3)(-154)} }{2(3)}$
x=$\frac{-19+/- \sqrt{361+1848} }{6}$
x=$\frac{-19+/- \sqrt{2209} }{6}$
x=$\frac{-19+/- 47 }{6}$

so
x=$\frac{-19+ 47 }{6}$ and x=$\frac{-19- 47 }{6}$
x=$\frac{28}{6}$ and x=$\frac{-66}{6}$
x=14/3 and -11