Given:
#y=-(x-3)^2-(2x-3)^2-10#
Expand #(x-3)^2# to #x^2-6x+9#.
#y=-(x^2-6x+9)-(2x-3)^2-10#
Expand #(2x-3)^2# to #4x^2-12x+9#.
#y=-(x^2-6x+9)-(4x^2-12x+9)-10#
Distribute the #-1# through both expressions.
#y=-x^2+6x-9-4x^2+12x-9-10#
Collect like terms.
#y=(-x^2-4x^2)+(6x+12x)+(-9-9-10)#
Simplify.
#y=-5x^2+18x-28# is a quadratic equation in standard form:
#y=ax^2+bx+c#,
where:
#a=-5#, #b=+18#, and #c=-28#
Find the roots using the quadratic formula.
The roots are the values for #x# when #y=0#, so first substitute #0# for #y#.
#0=-5x^2+18x-28#
Quadratic formula
#x=(-b+-sqrt(b^2-4ac))/(2*a)#
Plug in the known values.
#x=(-18+-sqrt(18^2-4*-5*-28))/(2*-5)#
Simplify.
#x=(-18+-sqrt(-236))/(-10)#
Prime factorize #-236#.
#x=(-18+-sqrt(-1xx2xx2xx59))/(-10)#
Simplify.
#x=(-18+-2isqrt59)/(-10)#
Two negatives make a positive.
#x=(18+-2isqrt59)/10#
Simplify by dividing by #2#.
#x=(color(red)cancel(color(black)(18))^9+-color(red)cancel(color(black)(2))^1isqrt59)/color(red)cancel(color(black)(10))^5#
#x=(9+-isqrt59)/5#
Solutions for #x# (the roots)
#x=(9+isqrt59)/5,##(9-isqrt59)/5#
They can also be written with the imaginary number #i# after the square root.
#x=(9+sqrt(59)i)/5,##(9-sqrt(59)i)/5#
