Question 346176
In answer to Your Question:
Solution Set would be the values found for {{{x[1]}}} , {{{x[2]}}} 

{{{x^2-3x=7}}}
Written as an Quadratic Equation:
{{{x^2-3x-7=0}}}
.
Using the Quadratic Formula to Solve:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
.
*[invoke quadratic "x", 1, -3, -7]