Verified answer

50y² - 50y - 28

Fist find the factors of 50*(-28) = -1*2*5*5*2*2*7

Now group them in 2 numbers so that their sum will be -50

50y² - 70y + 20y - 28

check 50*(-28) = -1400 and -70*20 = 1400

=10y(5y - 7) + 4(5y - 7)

=(5y - 7)(10y + 4)

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

: 50y^2 - 50y - 28

First reduce the coefficients to their simplest numbers.

2(25y^2 - 25y - 14)

Next write down all the factors of 25 (the coefficient of y^2)

25 x 1 = 25

5 x 5 = 25

Similarly for '14'

14 x 1 = 14

7 x 2 = 14

We then need to select a pair of factors of '25' and a pair from '14' which when multiplied together and then added/ssubtracted, equal '25' ( the coefficient of y)

Hence

5 x 5 & 7 x 2

5 x 7 = 35 & 5 x 2 = 10

35 - 10 = 25

placing inside brackets

(5y 2)(5y 7)

Next to select the signs.

Since '28' is negative, the two signs are different.

Since 50y is negative the larger multiple is negative and the other positive.

Hence

(5y + 2)(5y - 7)

Overall

2(5y + 2)(5y - 7) Fully factored!!!!

50y^2 - 50y - 28

=2(25y^2-25y-14)

solve 25y^2-25y-14

d=(-25)^2-4*25*(-14)

=2025

root d=root(2025)

root d=45

y=(-b+/-root d)/2a

y=(25+45)/50 or y=(25-45)/50

y=7/5 or y= -2/5

if p & q are 2 zeros of a quardratic equation, then the its factors are:

a(y-p)(y-q)

so factors are:

25(y-7/5)(y+2/5)

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

so, 50y^2-50a-28=2(5x-7)(5y+2)

Good luck !