Question 396732
Standard form for an equation like this is:
{{{ax^2 + bx + c = 0}}}
Note the zero on one side. So we start by getting one side of
{{{x^2+6x=3}}}
to be a zero. The easiest thing is to add -3 to each side:
{{{x^2 + 6x + (-3) = 0}}}
We can now "read" the values for a, b and c:
a = 1
b = 6
c = -3<br>
Solving the equation with the Quadratic Formula we get:
{{{x = -(6) +- sqrt((6)^2 - 4(1)(-3)))/2(1)}}}
which simplifies as follows:
{{{x = (-(6) +- sqrt(36 - 4(1)(-3)))/2(1)}}}
{{{x = (-(6) +- sqrt(36 + 12))/2(1)}}}
{{{x = (-(6) +- sqrt(48))/2(1)}}}
{{{x = (-6 +- sqrt(48))/2}}}
{{{x = (-6 +- sqrt(16*3))/2}}}
{{{x = (-6 +- sqrt(16)*sqrt(3))/2}}}
{{{x = (-6 +- 4*sqrt(3))/2}}}
{{{x = (2(-3 +- 2*sqrt(3)))/2}}}
{{{x = (cross(2)(-3 +- 2*sqrt(3)))/cross(2)}}}
{{{x = -3 +- 2*sqrt(3)}}}
In long form this is:
{{{x = -3 + 2*sqrt(3)}}} or {{{x = -3 - 2*sqrt(3)}}}
These are the simplified, exact expressions for the solutions to your equation.