I assume the equation is
[tex]\dfrac6{x^2+2x-15}+\dfrac7{x+5}=\dfrac2{x-3}[/tex]
Notice that
[tex]x^2+2x-15=(x+5)(x-3)[/tex]
so to get each fraction to have a common denominator, we need to rewrite
[tex]\dfrac7{x+5}=\dfrac{7(x-3)}{(x+5)(x-3)}=\dfrac{7x-21}{x^2+2x-15}[/tex]
and
[tex]\dfrac2{x-3}=\dfrac{2(x+5)}{(x+5)(x-3)}=\dfrac{2x+10}{x^2+2x-15}[/tex]
So we have
[tex]\dfrac6{x^2+2x-15}+\dfrac{7x-21}{x^2+2x-15}=\dfrac{2x+10}{x^2+2x-15}[/tex]
Combine the fractions and put them on one side:
[tex]\dfrac{6+(7x-21)-(2x+10)}{x^2+2x-15}=0[/tex]
If x ≠ -5 and x ≠ 3, we can ignore the denominator, leaving us with
[tex]6+(7x-21)-(2x+10)=0[/tex]
[tex](6-21-10)+(7x-2x)=0[/tex]
[tex]-25+5x=0[/tex]
[tex]5x=25[/tex]
[tex]x=\dfrac{25}5=\boxed{5}[/tex]
