Respuesta :

Billymkhoi
2k(7-5k)+11=6k+3(k^2-1) 
⇒ (2k)* 7 - (2k)* (5k) +11= 6k+ 3* k^2- 3*1 (distributive property)
⇒ 14k -10k^2 +11= 6k+ 3k^2 -3
⇒ (14k -14k) +(-10k^2 +10k^2)+ (11-11) = (6k-14k)+ (3k^2+10k^2)+ (-3-11)
⇒ 13k^2- 8k- 14= 0
⇒ 13 (k^2- 8/13k- 14/13)=0
⇒ (k- ((4+3*sqrt(22))/13))* ((k- ((4-3*sqrt(22))/13))= 0
⇒ k= (4+3*sqrt(22))/13 or k= (4-3*sqrt(22))/13).

Hope this helps~
cambeachlife

Answer:

k= (4+3*sqrt(22))/13 or k= (4-3*sqrt(22))/13).

Step-by-step explanation:

2k(7-5k)+11=6k+3(k^2-1)

We move all terms to the left:

2k(7-5k)+11-(6k+3(k^2-1))=0

We add all the numbers together, and all the variables

2k(-5k+7)-(6k+3(k^2-1))+11=0

We multiply parentheses

-10k^2+14k-(6k+3(k^2-1))+11=0

We calculate terms in parentheses: -(6k+3(k^2-1)), so:

6k+3(k^2-1)

We multiply parentheses

3k^2+6k-3

Back to the equation:

-(3k^2+6k-3)

We get rid of parentheses

-10k^2-3k^2+14k-6k+3+11=0

We add all the numbers together, and all the variables

-13k^2+8k+14=0

a = -13; b = 8; c = +14;

Δ = b2-4ac

Δ = 82-4·(-13)·14

Δ = 792

The delta value is higher than zero, so the equation has two solutions

We use following formulas to calculate our solutions:

k1=−b−Δ√2ak2=−b+Δ√2a

The end solution:

Δ−−√=792−−−√=36∗22−−−−−−√=36−−√∗22−−√=622−−√

k1=−b−Δ√2a=−(8)−622√2∗−13=−8−622√−26

k2=−b+Δ√2a=−(8)+622√2∗−13=−8+622√−26

   

   

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