Question

2a(5a+10)+(2a-8)(3a+2)=0

Answer 1

10а+20а+6а^2 + 4а - 24а - 16=0

6а^2 + 10а - 16=0

D=100+ 384=484

х1= -10-22\12=две целых две третьих

х2=-10+22\12=1

Answer 2

To solve the equation 2a(5a+10) + (2a-8)(3a+2) = 0, we'll start by simplifying the expression:

2a(5a+10) + (2a-8)(3a+2) = 0

First, let's simplify each term separately:

2a(5a+10) = 10a^2 + 20a

(2a-8)(3a+2) = 6a^2 + 4a - 24a - 16 = 6a^2 - 20a - 16

Now, substitute the simplified terms back into the equation:

10a^2 + 20a + 6a^2 - 20a - 16 = 0

Combining like terms:

16a^2 - 16 = 0

Divide both sides by 16:

a^2 - 1 = 0

Now, factor the quadratic equation:

(a-1)(a+1) = 0

So, the possible values of 'a' that satisfy the equation are a = 1 and a = -1.