Question

How do you solve `x(x-9)+20=0`?

Answer 1

Develop the equation

`x^2-9x+20=0`

Your equation is of the form

`ax^2-bx+c=0`

Find the discriminant (`Delta`)

`Delta = b^2-4ac`

`Delta = (-9)^2-4*20= 1`

`Delta` is greater than zero, this means two real roots exist.

`x'= (-b+sqrt(Delta))/(2a)= (-(-9)+ sqrt(1))/2 =5`
`x''= (-b-sqrt(Delta))/(2a)= (-(-9)- sqrt(1))/2 =4`