Question

5(a+2)+6a=25+4a(a+1)

Answer 1

Aplicando a distributiva

$5(a+2)+6a=25+4a(a+1)\\5a+10+6a=25+4a^2+4a\\11a+10=25+4a^2+4a$

$25+4a^2+4a=11a+10\\4a^2-7a+15=0$


$4a^2-7a+15=0\\\Delta=(-7)^2-4.4.15\\\Delta=-191$

$a= \frac{-b\pm \sqrt{\Delta} }{2a}$

$a= \frac{7\pm \sqrt{-191} }{8} ~\to~a= \frac{7\pm i\sqrt{191} }{8}$

$\boxed{a'= \frac{7+i \sqrt{191} }{8} }$

$\boxed{a''= \frac{7-i \sqrt{191} }{8} }$

Lembrando que :

$\sqrt{-1} =i$

$\boxed{\sqrt{-191} ~\to~ \sqrt{-1.191} ~\to~i \sqrt{191} }}$

Answer 2

5(a+2)+6a=25+4a(a+1)
5a +10 + 6a = 25 + 4a² + 4a 
 25 + 4a² + 4a - 5a -10 - 6a = 0
 4a² - 7a + 15 = 0

Δ= (-7)² - 4.4.15 = 49 - 256= - 207  raízes imaginárias.