a: =>(2x-5)(x-1)+2x(x+3)=4
=>2x^2-7x+5+2x^2+6x=4
=>4x^2-x+1=0
=>\(x\in\varnothing\)
b: =>\(\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{x^2-9}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{12x+42}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)
=>13x+39+x^2-9=12x+42
=>x^2+13x+30-12x-42=0
=>x^2+x-12=0
=>(x+4)(x-3)=0
=>x=-4