Answer:
The positive real number is 26
Step-by-step explanation:
Let
x ----> the number
we know that
The algebraic expression that represent this problem is
[tex]x^{2} =15x+286[/tex]
so
[tex]x^{2}-15x-286=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}-15x-286=0[/tex]
so
[tex]a=1\\b=-15\\c=-286[/tex]
substitute in the formula
[tex]x=\frac{-(-15)(+/-)\sqrt{-15^{2}-4(1)(-286)}} {2(1)}[/tex]
[tex]x=\frac{15(+/-)\sqrt{1,369}} {2}[/tex]
[tex]x=\frac{15(+/-)37}{2}[/tex]
[tex]x_1=\frac{15(+)37}{2}=26[/tex]
[tex]x_2=\frac{15(-)37}{2}=-11[/tex] ---> the solution cannot be negative
therefore
The positive real number is 26
