Answer:
70
Step-by-step explanation:
In parallelogram ABCD, diagonals AC and BD intersect at point E
E is the midpoint of BD
So BE = DE
Given: BE =2x^2-3x, and DE=x^2+10
BE = DE
[tex]2x^2-3x = x^2+10[/tex] (solve for x)
Subtract x^2 and 10 on both sides
[tex]2x^2-3x- x^2-10=0[/tex]
[tex]x^2-3x-10=0[/tex]
Factor x^2-3x-10
(x-5)(x+2) = 0
x-5 =0 so x=5
x+2=0 so x=-2
Length x cannot be negative so we ignore x=-2
Lets plug in 5 for x and find out BE and DE
BE =[tex]2x^2-3x= 2(5)^2-3(5)= 50-15= 35[/tex]
DE =[tex]x^2+10= (5)^2+10=35[/tex]
BD = BE + DE= 35+35 = 70
So BD = 70
