AS Maths help!!

Can someone help me solve the following equation? Please!
X^2 + 2px + (3p+4) = 0

Assuming that you mean $x^2 + 2px + (3p+4) = 0$, have you tried applying the quadratic formula?

Can u post a picture of the full question.

This question reminds me of a root of a quadratic question. Something along the lines of 'this quadratic has no real roots find the possible values of p'.

I could be wrong tho

The equation x^2 + 2px + (3p+4) = 0, where p is a positive constant, has equal roots. Find the value of p and solve the equation.

I'm no sure as to whether this is the definite answer, I might've gone wrong somewhere along the lines but here's my try:

(Everything in [square brackets] are intended to be written as fractions to avoid confusion)

x²+2px+(3p+4)=0
x²+4=-2px-3p
x²+4=p(-2x-3)
p=[x²+4/-2x-3]

x²+2([x²+4/-2x-3])x+3([x²+4/-2x-3])
x²+[2x²+8/-4x-6]x+[3x²+12/-6x-9]+4=0

x²+[2x²+8/-2(2x+3)]x-[3(x²+4)/3(2x+3)]+4=0
x²+[2x³+8x/-2(2x+3)]-[3(x²+4)/3(2x+3)]+4=0
x²-[2(x³+4x)/2(2x+3)]-[(x²+4)/(2x+3)]+4=0

[x²/1]-[x³+4x/2x+3]-[x²+4/2x+3]+[4/1]=0
[x²x(2x+3)-(x³+4x)-(x²+4)+4(2x+3)/2x+3]

[2x³+3x²-x³-4x-x²-4+8x+12/2x+3]

x³+2x²+4x+8=0

(x²+4)(x+2)=0
x²+4=0
x+2=0

x∉ℝ
x=-2

Apologies if it isn't correct!

As i suspected.

Equal roots means b^2-4ac=0, so u can work out p, then substitute back in and solve

Ty! Next time I'll bare in mind,
equal roots = discriminant