Question 304624
find the real solutions, using the quadratic equation
3x(x+2)=1
Convert equation to the standard form: ax^2 + bx + c = 0
3x^2 + 6x - 1 = 0
The quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
in this problem: a=3; b=6; c=-1
{{{x = (-6 +- sqrt( -6^2 - 4*3*-1 ))/(2*3) }}}
:
{{{x = (-6 +- sqrt(36 - (-12) ))/6 }}}
:
{{{x = (-6 +- sqrt(36 + 12 ))/6 }}}
:
{{{x = (-6 +- sqrt(48 ))/6 }}}
Two solutions
{{{x = (-6 + 6.928)/6 }}}
x = {{{.928/6}}}
x = .15467
and
{{{x = (-6 - 6.928)/6 }}}
x = {{{(-12.928)/6}}}
x = -2.15467