The equation 3 x ^ 2 + x + 5 = x - 3 , where x is real, has

\sqrt{3{x}^{2} + x + 5} = x - 3 . . . \left(1\right)

We know that$for a quadratic expressiona{x}^{2} + bx + c, a > 0,$ if discriminant i.e. $D = {b}^{2} - 4ac < 0,$ then the quadratic is always positive for all real values of the variable.

Now, consider $3{x}^{2} + x + 5$

We have, discriminant $= {b}^{2} - 4ac = 1 - 4 \times 3 \times 5 < 0$

$∴ 3{x}^{2} + x + 5 > 0, \forall x \in R$

Squaring \left(1\right),$we get,</p><p>$3{x}^{2} + x + 5 = {\left(x - 3\right)}^{2}

$⇒ 3{x}^{2} + x + 5 = {x}^{2} - 6x + 9$

$⇒ 2{x}^{2} + 7x - 4 = 0$

$⇒ \left(x + 4\right)\left(2x - 1\right)$ = 0$</p><p>$⇒ x = - 4,x = $\dfrac{1}{2}$

But $x - 3 < 0$ for $x = - 4, \dfrac{1}{2}$

Hence, the equation has no solution.