Question 264878
equation is:


x^2-10x= -9 


add 9 to both sides of this equation to get it into standard form.


you get:


x^2 - 10x + 9 = 0


standard form of the quadratic equation is ax^2 + bx + c = 0


this means that:


a = 1
b = -10
c = 9


quadratic formula is:


x = (-b +- sqrt(b^2-4ac))/(2a)


this becomes:


(-(-10) +- sqrt((-10)^2 - 4*1*9)/(2*1)


this becomes:


(10 +- sqrt(100-36))/2


this becomes:


(10 +- sqrt(64))/2


this becomes:


(10 +- 8)/2


this becomes:


18/2 = 9

or:

2/2 = 1


you get:


x = 9 or x = 1


substitute in original equation to see if these values are good.


original equation is:


x^2 - 10x = -9


when x = 9, this becomes:


81 - 90 = -9 which becomes -9 = -9 which is true.


when x = 1, this becomes:


1 - 10 = -9 = -9 which is also true.


the values are confirmed to be good.


your solutions are:


x = 9 and x = 1


these are the real roots of this equation which means that the graph of the equation crosses the x-axis at these points.


a graph of your equation looks like this:


{{{graph (600,600,-10,10,-20,20,x^2 - 10x + 9)}}}