Answer:
[tex]\large\boxed{u=-\dfrac{2}{5}\ or\ u=3}[/tex]
Step-by-step explanation:
[tex]5u^2=13u+6\qquad\text{subtract}\ 13u\ \text{and 6 from both sides}\\\\5u^2-13u-6=0\\\\5u^2-15u+2u-6=0\qquad\text{distribute}\\\\5u(u-3)+2(u-3)=0\\\\(5u+2)(u-3)=0\iff5u+2=0\ \vee\ u-3=0\\\\5u+2=0\qquad\text{subtract 2 from both sides}\\\\5u=-2\qquad\text{divide both sides by 5}\\\\\boxed{u=-\dfrac{2}{5}}\\\\u-3=0\qquad\text{add 3 to both sides}\\\\\boxed{u=3}[/tex]
