Answer:
-4 and 9
Step-by-step explanation:
we have
[tex]x^{2} -36=5x[/tex]
we know that
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} -36=5x[/tex]
[tex]x^{2}-5x-36=0[/tex]
so
[tex]a=1\\b=-5\\c=-36[/tex]
substitute in the formula
[tex]x=\frac{-(-5)(+/-)\sqrt{-5^{2}-4(1)(-36)}} {2(1)}[/tex]
[tex]x=\frac{5(+/-)\sqrt{169}} {2}[/tex]
[tex]x=\frac{5(+/-)13} {2}[/tex]
[tex]x=\frac{5(+)13} {2}=9[/tex]
[tex]x=\frac{5(-)13} {2}=-4[/tex]
therefore
The equation is true for -4 and 9
