3x2+3x-15=\(\sqrt{x^2+x+3}\)
=>3x2+3x+9-24=\(\sqrt{x^2+x+3}\)
=>3(x2+x+3)-24=\(\sqrt{x^2+x+3}\) (1)
đặt \(\sqrt{x^2+x+3}=t\) >.hoặc =0
(1)<=>3t2-t-24=0
<=>3t2-9t+8t-24=0
<=>3t(t-3)+8(t-3)=0
<=>(t-3)(3t+8)=0
=>t=3
hoặc t=\(\dfrac{-8}{3}\)(loại)
=>\(\sqrt{x^2+x+3}=3\)
=>x2+x+3=9
=>x2+x-6=0
=>x2+3x-2x-6=0
=>x(x+3)-2(x+3)=0
=>(x+3)(x-2)=0
=>x=-3 hoặc x=2