Question

$24y + 35 + 4y {}^{2}$

Answer 1

We can write

24y + 35 + 4y² as

4y² + 24y + 35

Now this is in the form of ax² + bx + c = 0
So we can split the middle term,

we get

4y² + 10y + 14y + 35 = 0

=> 2y(2y +5) + 7(2y + 5) = 0

=> (2y + 5)(2y + 7) = 0


Hence, factorisation = (2y + 5)(2y + 7)

If you want to solve for y then :-

(2y + 5)(2y + 7) = 0

=> (2y + 5) = 0/(2y + 7)

=> 2y + 5 = 0

=> 2y = -5

=> y = -5/2


Or,

(2y + 5)(2y + 7) = 0

=> 2y + 7 = 0/(2y + 5)

=> 2y + 7 = 0

=> 2y = -7

=> y = -7/2


Hope it helps dear friend ☺️✌️

Answer 2

$24y + 35 + 4y {}^{2} = 4y {}^{2} + 24y + 35 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4y {}^{2} + (14 + 10)y + 35 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4y {}^{2} + 14y + 10y + 35 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2y(2y + 7) + 5(2y + 7) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (2y + 5)(2y + 7)$






If you want to find the value of "y"...Then


4y^2+24y+35=0

=>(2y+5)(2y+7)=0


So,

2y+5=0

=>2y=-5

=>y=-5/2


Or,

2y+7=0

=>2y=-7

=>y=-7/2


Hence,

The values of y = -5/2 , -7/2