How to solve this quadratic congruence? $27w^2+20w+35 \equiv 0 \pmod{23}$ Given $27w^2+20w+35 \equiv  0 \pmod{23}$
How to solve this  quadratic congruence?
All hints are welcome.
I completed the square ,so I obtained $27(x+\frac{10}{27})^2+\frac{845}{27} \equiv  0 \pmod{23} $
I need to get 2 solutions in the least residue system which are the correct answers:  $w\equiv  6 \pmod{23}$ and  $w \equiv  12 \pmod{23}$
 A: For ease of manual computation we rewrite the equation as
$$4w^2-3w+12\equiv0\bmod23$$
Divide by the leading coefficient, i.e. multiply by $4^{-1}=6$:
$$w^2+5w+3\equiv0\bmod23$$
Now apply the quadratic formula:
$$w\equiv\frac{-5\pm\sqrt{13}}2\bmod23$$
We need to work out the square roots of $13$ in $\mathbb Z_{23}$. $6$ is easily verified as one root, so $-6$ is the other:
$$w\equiv\frac{-5\pm6}2\equiv9\pm3\bmod23$$
A: Hint:
$$\pmod{23}: 4w^2-3w+12\equiv 0 \implies 8w^2-6w+1\equiv 0  \implies (2w-1)(4w-1)\equiv 0. $$
Update To justify why I multiply 2 to $4w^2-3w+12$, it's easier to work with whole numbers than fractions, so to complete square while keeping every coefficients integer we multiply by 16:
$$16(4w^2-3w+12)=64w^2-48w+192=(8w-3)^2+183\equiv (8w-3)^2-1 = (8w-2)(8w-4)=8(4w-1)(2w-1) \pmod{23}$$
and now you see why.
Update 2: I like Parcly Taxel's way of making the quadratic monic first:
$$w^2+5w+3\equiv0\pmod{23}$$
After that it can be done a little faster:
$$w^2-18w+3\equiv 0 \implies (w-9)^2 = 78\equiv 9 =3^2 \implies (w-6)(w-12) \equiv 0 \pmod{23}$$
A: Since $27 \equiv 4$ we can write the equation as $4w^2 + 20w + 35 \equiv 0.$ Completing the square gives $(2w+5)^2 + 10 \equiv 0,$ i.e. $(2w+5)^2 \equiv -10.$ But $-10 \equiv -10+2\cdot 23=36=6^2,$ so $2w+5\equiv\pm 6,$ i.e. $2w=-5\pm 6.$
Case $+$: $2w=-5+6=1\equiv 1+23=24=2\cdot12$ so $w\equiv12.$
Case $-$: $2w=-5-6=-11\equiv -11+23=12=2\cdot6$ so $w\equiv6.$
Thus the solutions are $w=12$ and $w=6$.
