If y=mx+3 is a tangent to the parabola y=-2x^2+3x-5,then m=?

If y=mx+3 is a tangent to the parabola y=-2x^2+3x-5, there must be one and only one solution to this set of equations.

mx+3 = -2x^2+3x-5
2x^2 + (m-3)x + 8 = 0

for this equation having only one solution, its discriminant ( b^2 - 4ac) = 0,

i.e. (m-3)^2 - 4(2)(8) = 0

(m-3)^2 - 8^2 = 0
(m-3+8)(m-3-8)=0 [difference of two squares)
(m+5)(m-11)=0

thus m = -5 or 11

y =-2x^2 + 3x - 5
y= mx + 3
y= -2x^2 + 3x - 5 = mx + 3
-2x^2 + ( 3 - m ) x - 5 - 3 = 0
2x^2 - ( 3 - m ) x + 8 =0
△ = 0 ( has only one root )
△ = [ - ( 3 - m ) ]^2 - 4 (2)(8) = 0
9 + m^2 - 6m - 64 = 0
m^2 - 6m - 55 = 0
( m - 11 )( m + 5 ) = 0
m = 11 or -5