+1491 I'm going on the assumption that "x2" means \({x}^{2}\)
We start with:
28=x^2-3x
If you want to find solutions then you need to have the equation equal zero:
0=x^2-3x-28
If you factor then you get:
0=(x+4)(x-7)
Solve each binomial as if it were a separate problem
x+4=0 and x-7=0
Meaning...
x=-4 and 7
+1491 I'm going on the assumption that "x2" means \({x}^{2}\)
We start with:
28=x^2-3x
If you want to find solutions then you need to have the equation equal zero:
0=x^2-3x-28
If you factor then you get:
0=(x+4)(x-7)
Solve each binomial as if it were a separate problem
x+4=0 and x-7=0
Meaning...
x=-4 and 7
