How do you solve #x^2 – 6x -7 = 0#? Algebra Quadratic Equations and Functions Quadratic Functions and Their Graphs 1 Answer Konstantinos Michailidis Mar 16, 2016 We can write this as follows #x^2-6x-6-1=(x^2-1)-6(x+1)=(x-1)(x+1)-6(x+1)= (x+1)*(x-1-6)=(x+1)*(x-7)# Now from #x^2-6x-7=0=>(x+1)*(x-7)=0# we get that the roots are #x=-1#, #x=7# Answer link Related questions What are the important features of the graphs of quadratic functions? What do quadratic function graphs look like? How do you find the x intercepts of a quadratic function? How do you determine the vertex and direction when given a quadratic function? How do you determine the range of a quadratic function? What is the domain of quadratic functions? How do you find the maximum or minimum of quadratic functions? How do you graph #y=x^2-2x+3#? How do you know if #y=16-4x^2# opens up or down? How do you find the x-coordinate of the vertex for the graph #4x^2+16x+12=0#? See all questions in Quadratic Functions and Their Graphs Impact of this question 863 views around the world You can reuse this answer Creative Commons License