Question 423723
let x^2=y We have to find y and then x...
The general formula for {{{ay^2+by+c=0}}} is {{{y=(-b+-sqrt(b^2-4ac))/(2a)}}}
Where a=10,b=-19, and c=6
{{{y=(-(-19)+-sqrt((-19)^2-4*10*6))/(2*10)}}}
{{{y=(19+-sqrt(361-240))/20}}}=
{{{(19+-sqrt(121))/20}}}=
{{{(19+-sqrt(11^2))/20}}}=

We obtain two values for y:
y=3/2 or y=8/20=2/5=0.4
Each one of them gives two values for x:
x={{{0+-sqrt(y)}}}
x={{{sqrt(3/2)))}}}
or
x=-{{{sqrt(3/2)))}}}
or
x={{{sqrt(0.4)}}}
or
x=-{{{sqrt(0.4)}}}
So we have four solutions:

{{{sqrt(3/2)))}}};-{{{sqrt(3/2)))}}};{{{sqrt(0.4)}}};-{{{sqrt(0.4)}}}