Solve the congruence

Question

Solve the congruence .

$3x^2+7x+5=0 \pmod{11}$

When I m trying to solve this question I.e. Multiplying by $4a$

$36x^2+84x+60=0 \pmod{11}$

$(2ax+b)^2=(b^2-4ac)$

AND So we got

$(6x+7)^2=-11 \pmod{11}$

$y^2=-11\pmod{11}$

I don't know how to solve it from here.

Answer 1

$$(6x+7)^2\equiv-11\pmod{11}\equiv0$$

$\iff6x+7\equiv0\iff6x\equiv-7\iff12x\equiv-14$ as $(2,11)=1$

$\iff x\equiv-14\equiv8$

Answer 2

$$3x^2 +7x + 5 \equiv 0 \mod 11$$

• Aiming at square completion, note that multiplying by $7$ is useful: $$\Leftrightarrow 21x^3+49x+35 \equiv -x^2 + 5x + 2 \equiv 0 \mod 11 $$ $$\Leftrightarrow x^2 -5x - 2 \equiv 0 \mod 11$$ $$\stackrel{11-5=6, 11-2=9}{\Longleftrightarrow} x^2 + 6x +9 \equiv (x+3)^2 \equiv 0 \mod 11$$ It follows: $$\boxed{x \equiv -3 \equiv 8 \mod 11}$$