Rom

let the first integer be n

the second is (n+1), the third (n+2)

4n = (n+2) + 13

4n = n + 15

3n = 15

n = 5

4 x 5 = 20 = 7 + 13

the 3 integers are 5, 6, 7

did you try dividing? no.. as you clearly tried nothing 

$5x^2 + 10x - 15 = (x+5) 5x -(15x+15) \\ -(15x+15) = -15(x+5) +60 \\ 5x^2 + 10x - 15 = (15x-15)(x+5) + 60 \\ \text{The remainder is 60}$

we can add the two equations to obtain

$2x^2 +(a+c)x+(b+d)=0\\ x^2 + \dfrac{a+c}{2}x+\dfrac{b+d}{2}=0$

$\text{Now just apply the quadratic formula}\\ r_{1,2} = \dfrac{-\frac{a+c}{2}\pm \sqrt{\left(\frac{a+c}{2}\right)^2 - 2(b+d)}}{2}$

$\text{we can clean this up a bit}\\ r_{1,2} = \frac{1}{4} \left(\pm\sqrt{(a+c)^2-8 (b+d)}-a-c\right)$

$\text{If there is only a single shared root }r_1 = r_2\\ \text{if there are no shared roots}\\ (a+c)^2 - 8(b+d)<0$

Px + 40 = Qx + 20

Px +20 = Qx

20 = Qx - Px = (Q-P)x

If Q=P then (Q-P)x = 0 for all x and thus the original equation has no solutions.

I'm only seeing 3 triplets of integers that meet the specifications

(1, 3, 8), (1, 4, 5), (2, 2, 4)

Thus the sum asked for is

8 + 5 + 4 = 17

you got the first one right

for (b)

3x = 15

divide both sides by 3 to get

x=5

for (c)

x = 12-8 = 4

$Px - 37 = Qx - 37 \\ Px = Qx \\ P=Q \text{ leads to infinitely many solutions} \\ P\neq Q \text{ leads to the single solution } x=0$

$x^4-11x^3+24x^2=x^2(x^2-11x+24) \\ (4x^2-44x+96) = 4(x^2-11x+24) \\ x^4-11x^3+24x^2 - (4x^2-44x+96) =(x^2-11x+24)(x^2-4) = \\ (x-8)(x-3)(x-2)(x+2) \\ \text{and thus the solutions are }\\ x=8,~3,~2,~-2$

ever think that B might be female?

These angles need to sum to 360 degrees.

120 + 80 + B = 360

B = 160 degrees