Question 238974
 use the quadratic formula to solve the equation.
{{{x^2-5x=-9}}}
The solution set is 
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The quadratic formula to solve the equation

{{{Ax^2+Bx+C=0}}} is {{{x = (-B +- sqrt(B^2-4AC ))/(2A) }}} 

Get your equation in the form {{{Ax^2+Bx+C=0}}} by adding 9
to both sides:

{{{x^2-5x=-9}}}

{{{x^2-5x+9=0}}}

Then {{{A=1}}}, {{{B=-5}}}, and {{{C=9}}}

Substitute those into:

{{{x = (-B +- sqrt(B^2-4AC ))/(2A) }}}

{{{x = (-(-5) +- sqrt((-5)^2-4(1)(9) ))/(2(1)) }}}

{{{x = (5 +- sqrt(25-36 ))/2 }}}

{{{x = (5 +- sqrt(-11))/2 }}}

{{{x = (5 +- i*sqrt(11))/2 }}}

The solution set is:

{{{"{"}}}{{{(5 + i*sqrt(11))/2}}}{{{","}}}{{{(5 - i*sqrt(11))/2}}}{{{"}"}}}
 
Edwin</pre>