How do you find the roots, real and imaginary, of y=-(x -4 )^2-2x-1 using the quadratic formula? Algebra Quadratic Equations and Functions Quadratic Formula 1 Answer Tamir E. Dec 27, 2017 x_1=3+2sqrt2i x_2=3-2sqrt2i Explanation: y=-(x-4)^2-2x-1= =-(x^2-8x+16)-2x-1= =-x^2+8x-16-2x-1= =-x^2+6x-17 Let y=0 => -x^2+6x-17=0 iff a=-1, b=6, c=-17 => x_{1,2}={-b+-sqrt{b^2-4ac}}/{2a}= ={-6+-sqrt{6^2-4*(-1)*(-17)}}/{2(-1)}= ={-6+-sqrt{36-68}}/{-2}= ={-6+-sqrt{-32}}/{-2}= ={-6+-sqrt{-(16*2)}}/{-2}= ={-6+-4sqrt2i}/{-2}= ={cancel(-2)(3+-2sqrt2i)}/{cancel(-2)}= =3+-2sqrt2i => x_1=3+2sqrt2i x_2=3-2sqrt2i Answer link Related questions How do you know how many solutions 2x^2+5x-7=0 has? What is the Quadratic Formula? How do you derive the quadratic formula? How is quadratic formula used in everyday life? How do you simplify the quadratic formula? How do you solve x^2+10x+9=0 using the quadratic formula? How do you solve -4x^2+x+1=0 using the quadratic formula? When do you have "no solution" when solving quadratic equations using the quadratic formula? How do you solve 4x^2=0 using the quadratic formula? Why can every quadratic equation be solved by using the quadratic formula? See all questions in Quadratic Formula Impact of this question 1521 views around the world You can reuse this answer Creative Commons License