We are given
#color(red)(y = - (x-1)^2 - x^2 - 3x+2)#
We need to use the quadratic formula to find the roots, real or imaginary.
The algebraic expression on the Right-hand-Side(RHS) can be simplified to ...
#color(blue)(-(x^2 - 2x +1) - x^2 - 3x + 2)#
#color(blue)(rArr -x^2 + 2x -1 - x^2 - 3x +2)#
#color(blue)(rArr -2x^2 -x +1 )#
Now we have
#y = -(2x^2 + x - 1)#
The General Form of a Quadratic Equation is
#color(red)(a)x^2 + color(red)(b)x + color(red)(c) = 0#
The Quadratic Formula for finding the roots of a quadratic equation is given by
#color(blue)(x=((-b)+- sqrt(b^2 - 4ac))/(2a),a!=0)#
#color(green)(b^2 - 4ac)# is referred to as the Discriminant
When #color(green)((b^2 - 4ac)=0)# there exists One Real Root
When #color(green)((b^2 - 4ac)>0)# there exists Two Real Roots
When #color(green)((b^2 - 4ac)<0)# there exists Two Complex Roots
For our Quadratic Equation
#-2x^2 - x + 1 = 0#
We observe that
#color(red)(a) = -2#
#color(red)(b) = -1#
#color(red)(c) = 1#
Substitute these values in the Quadratic Formula
#color(blue)(x=((-b)+- sqrt(b^2 - 4ac))/(2a)#
We get
#x=((-(-1))+- sqrt((-1)^2 - 4(-2)(1)))/(2(-2))#
On simplification we get,
#x=((+1)+- sqrt(+1 +8))/(-4)#
#x=(1+- sqrt(9))/(-4)#
We observe that, the discriminant #(b^2 - 4ac) > 0#
Hence, we will have Two Real Roots
#x = (1 +- 3)/-4#
#rArr x= (1+3)/-4; x = (1-3)/-4#
#rArr x= 4/-4; x = -2/-4#
#rArr x= -1; x = 1/2#
Hence our two real roots are
#color(blue)(x = 1/2 or x = 0.5)#
#color(blue)(x = -1)#
Hope this helps.
