Quadratic Equations

Given, equation is
√3x2+x+5=x−3
Here, 3x2+x+5≥0 (since square root is non-negative)
‌⇒‌‌x−3≥0
‌⇒‌‌x≥3
Now, squaring to both sides of given equation, we get
‌(√3x2+x+5)2=(x−3)2
⇒3x2+x+5=x2−6x+9
⇒2x2+7x−4=0
Using quadratic formula,
x‌=‌

−7±√72−4⋅2⋅(−4)

2⋅2

‌=‌

−7±√49+32

4

=‌

−7±9

4

⇒‌‌x‌=‌

2

4

‌ and ‌x=‌

−16

4

⇒‌‌x‌=‌

1

2

‌ and ‌x=−4
Since domain, x≥3, so x=−4
And x=‌

1

2

are not possible.
∴ There is zero solution.