Question 21800
well first clear the fraction <br>

(x^2+7x)/3 = -4
3((x^2+7x)/3 = -4)
x^2+4x=-12<br>

now simplify and solve using the quadratic formula but you will find there is no solution<br>

x^2+4x+12=0<br>

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} <br>

where a = 1, b = 4, & c = 12<br>

{{{x = (-4 +- sqrt( 4^2-4*1*12 ))/(2*1) }}}
{{{x = (-4 +- sqrt(16-4*12))/(2*1)}}}
{{{x = (-4 +- sqrt(16-48))/(2)}}}
{{{x = (-4 +- sqrt(-32))/(2)}}}<br>

since you can't take the square root of a negative number and get a real number back there are no real solutions.