3x2+x+5=x−3...(1)
We know that for a quadratic expression ax2+bx+c,a>0, if discriminant i.e. D=b2−4ac<0, then the quadratic is always positive for all real values of the variable.
Now, consider 3x2+x+5
We have, discriminant =b2−4ac=1−4×3×5<0
∴3x2+x+5>0,∀x∈R
Squaring (1), we get,
3x2+x+5=(x−3)2
⇒3x2+x+5=x2−6x+9
⇒2x2+7x−4=0
⇒(x+4)(2x−1)=0
⇒x=−4,x=21
But x−3<0 for x=−4,21
Hence, the equation has no solution.