Question

solve 5(3y+1)square +6(3y+1)-8=0​

Answer 1

Answer:

=>5(3y+1)²+6(3y+1)-8=0

First we will find the value of 5(3y+1)²

So,

5(3y+1)²=5[(3y)²+(2×3y×1)+(1)²]. I identity [(a+b)²=a²+2ab+b²]

=5[9y²+6y+1]

=45y²+30y+5

Now we will find,

5(3y+1)²+6(3y+1)-8=0

So,

(45y²+30y+5)+(18y+6)-8=0

45y²+30y+5+18y+6-8=0

45y²+48y-3=0

So here's your answer dear friend.

But you can still continue with a step....

i.e.,

45y²+48y=0+3

45y²+48y=3

Hope this will help you dear friend.

Answer 2

Answer:

Ya she's correct frnd...

Be happy...

Step-by-step explanation:

Hope it helps you