15x²+16x+4 =0 (ax² +bx +c=0)
Δ = b²-4ac =256 - 4×15×4 =16
x1 = (-b+√Δ) / 2a = (-16+√16) / 30 =( -16+4) / 30 = -12/30 = - 2/5
x2 = (-b -√Δ) / 2a = (-16 -√16) / 30 = (-16 -4) /30 = -20/30 = -2/3
First of all, we can observe that

So the expression becomes

This means that the expression is defined for every 
Now, since the denominator is always positive (when it exists), the fraction can only be positive if the denominator is also positive: we must ask

Since we can't accept 3 as an answer, the actual solution set is
![(-\infty,-2] \cup [2,3) \cup (3,\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C-2%5D%20%5Ccup%20%5B2%2C3%29%20%5Ccup%20%283%2C%5Cinfty%29)
Answer:
∠EFH = 21°
∠HFG = 62°
∠EFG = 83°
Step-by-step explanation:
The diagram showing the angles has been attached to this response.
From the diagram, it can be deduced that;
Angle EFG = angle EFH + angle HFG
=> ∠EFG = ∠EFH + ∠HFG -------------------(i)
From the question:
∠EFH = (5x + 1)° -------------(ii)
∠HFG = 62° -------------(iii)
∠EFG = (18x + 11)° -------------(iv)
<em>Substitute these values into equation (i) as follows;</em>
(18x + 11) = (5x + 1) + 62
=> 18x + 11 = 5x + 1 + 62
<em>Collect like terms and solve for x</em>
18x - 5x = 1 + 62 - 11
13x = 52
x = 4
Now, to get each measurement, substitute x = 4 into each of equations (ii) - (iv)
∠EFH = (5x + 1)°
∠EFH = (5(4) + 1)°
∠EFH = (20 + 1)°
∠EFH = 21°
∠HFG = 62° [<em>Does not depend on x</em>]
∠EFG = (18x + 11)°
∠EFG = (18(4) + 11)°
∠EFG = (72 + 11)°
∠EFG = 83°
<u>Conclusion:</u>
∠EFH = 21°
∠HFG = 62°
∠EFG = 83°