Question

1X^(2) + 28 - 48 = 0

Answer 1

1x^2-20=0
X^2=20
x=square root of 20 though i dont know if your question accepts decimals or not

Answer 2

if the question is actually
\[x ^{2} + 2x - 48\]
You need factors that when added equal 2 and when multiplied equal 48

factors of 48:
1x48
2x24
4x12
6x8

Will any of those work?

Answer 3

If the problem is x^2+28x-48, then there are no factors of -48 that will give the middle term of +28x. So, you need to use the Quadratic Formula to solve this problem.

The Quadratic Formula is: x = [-b plus or minus (square root of b^2 - 4 (a)(c))]/2a

Then:
a=1
b=28
c=-48

x = [-28 plus or minus (square root of (28)^2 - 4(1)(-48))]/2(1)
= [-28 plus or minus (square root of 784 + 192)]/2
= [-28 plus or minus (square root of 976)]/2
= [-28 plus or minus (square roots of (16)(61))/2
= [-28 plus or minus 4(square root 61)]/2
x = -14+2(square root 61) and x = -14-2(square root 61)