Solving 2d² – 13d + 8 = -8?

2d² – 13d + 8 = -8

Add 8 to both sides, because the equation must equal zero
2d² – 13d + 16 = 0

I don't think it can be factored, so I'll use the quadratic formula. If I missed something and I'm wrong, it'll still work
x = [-b ± √(b² - 4ac)] / 2a
x = [13 ± √(169 - 128)] / 4
x = [13 ± √(41)] / 4
x = 13/4 ± √41/4

Leave it at that, or...
x = 4.850781059 or 1.649218941
x ≈ 4.85 or 1.65

Edit: And d = x. I used x out of habit.

2d² - 13d + 8 = - 8
2d² - 13d = - 16
d² - 13/2d = - 8
d² - 13/4d = - 8 + (- 13/4)²
d² - 13/4d = - 128/16 + 169/16
(d + 13/4)² = 41/16
d + 13/4 = 1.600781

Values of d:
d = 1.600781 - 3.25, d = - 1.649219
d = - 1.600781 - 3.25, d = - 4.850781

Answer: d = - 1.649219, - 4.850781

2d^2 - 13d + 8 = -8
2d^2 - 13d + 8 + 8 = 0
2d^2 - 13d + 16 = 0
d = [-b ±√(b^2 - 4ac)]/2a

a = 2
b = -13
c = 16

d = [13 ±√(169 - 128)]/4
d = [13 ±√41]/4

∴ d = [13 ±√41]/4

2d² - 13d + 16 = 0
d = [ 13 ± √ (169 - 128 ) ] / 4
d = [ 13 ± √ (41) ] / 4

use quadratic formula

d1 = (-b + (b^2-4ac)^1/2)/2a
d2 = (-b + (b^2-4ac)^1/2)/2a

d1 = (13 + (41)^1/2)/4
d1 = (13 - (41)^1/2)/4

2d² – 13d + 8 = -8
2d² – 13d + 16 = 0

Using: if ad^2 +bd+c = 0 then
d= [-b ± √(b^2-4ac)]/2a

d = {13 ± √[(-13)^2-4*2*16)]}/(2*2)

d ≈ 4.85 or d ≈ 1.65

Hope this will help.

Regards,

Marcellus Tanjaya
email: [email protected]
[email protected]

2d^2 - 13d + 8 + 8 = -8 +8
2d^2 - 13d + 16 = 0

now you need two number with product of 32 and sum of -13 but i cant think of any numbers that will fit the criteria, so you have to use Quadratic formula to solve it ... just google quadratic formula, and follow the instructions on how to solve it

Every quadratic equation has two roots.

2d^2 - 13d + 8.8 = 0

Use the Quadratic formula now.

One of the two roots is 5.732
The other is for you to figure ...

Commit that formula to memory.