(1) 5x^2+2x-15=0 I suspect that this should be 5x^2+22x-15=0, because the equation as written does not factorise. Write down the factors of the squared term coefficient and the constant term. We write these as ordered pairs (a,b) squared term (c,d): (1,5) and constant: (1,15), (15,1), (3,5), (5,3). The sign of the constant (minus) tells us that we are going to subtract the cross-products of factors. If it had been plus we would be adding the cross-products. Now we create a little table:
Quadratic factor table
a |
b |
c |
d |
ad |
bc |
|ad-bc| |
1 |
5 |
15 |
1 |
1 |
75 |
74 |
1 |
5 |
1 |
15 |
15 |
5 |
10 |
1 |
5 |
3 |
5 |
5 |
15 |
10 |
1 |
5 |
5 |
3 |
3 |
25 |
22 |
The column |ad-bc| just means the difference between ad and bc, regardless of it being positive or negative, just take the smaller product away from the larger.
If the coefficient of the x term is in the last column, then that row contains the factors you need. If it isn't in the last column, then the quadratic doesn't factorise, or you've missed some factors.
If 22x is the middle term, then the factors are shown in the last row and (a,b,c,d)=(1,5,5,3). Now we look at the sign of the middle term. Whatever the sign is we put it in front of the number c or d for the larger cross-product. The cross-products are bc and ad. In this case, bc is bigger than ad so the + sign goes in front of c. The sign in front of the constant tells us whether the sign in front of d is different or the same. If the sign is plus it's the same, otherwise it's the opposite sign. So in this case, it's minus, so the minus sign goes in front of d and we have (ax+c)(bx-d) (note the order of the letters!) or (x+5)(5x-3), putting in the values for a, b, c and d. If (x+5)(5x-3)=0, then x+5=0 or 5x-3=0. So in the first case x=-5 and in the second case 5x=3 so x=3/5.
(2) 21x^2-25x-4=0 (moving 4 over to the left to put the equation into standard form).
Quadratic factor table
a |
b |
c |
d |
ad |
bc |
|ad-bc| |
1 |
21 |
1 |
4 |
4 |
21 |
17 |
1 |
21 |
4 |
1 |
1 |
84 |
83 |
1 |
21 |
2 |
2 |
2 |
42 |
40 |
3 |
7 |
1 |
4 |
12 |
7 |
5 |
3 |
7 |
4 |
1 |
3 |
28 |
25 |
3 |
7 |
2 |
2 |
6 |
14 |
8 |
The row in bold print applies. The sign in front of 4 is minus so the signs in the brackets will be different. 28 is the larger product, so since we have -25x, the minus sign goes in front of c (4) and plus in front of d (1): (3x-4)(7x+1)=0.
So the solution is x=4/3 or -1/7.
(3) 10x^2+3x-4=0.
Quadratic factor table
a |
b |
c |
d |
ad |
bc |
|ad-bc| |
1 |
10 |
1 |
4 |
4 |
10 |
6 |
1 |
10 |
4 |
1 |
1 |
40 |
39 |
1 |
10 |
2 |
2 |
2 |
20 |
18 |
2 |
5 |
1 |
4 |
8 |
5 |
3 |
2 |
5 |
4 |
1 |
2 |
20 |
18 |
2 |
5 |
2 |
2 |
4 |
10 |
6 |
(2x-1)(5x+4)=0, so x=1/2 or -4/5.