I have tried the quadratic formula and also completing the square(sorta).

(questions are not in standard form yet)

6x2 = 4 - 5x

4x2 + 2x = 5

9(x - 4)2 + 1 = 0

Hello, Cameron!

I have tried the Quadratic Formula and also completing the square (sorta).
If you showed some work, we might find your errors.

(questions are not in standard form yet)
Did you put them in standard form?

$[1]\;6x^2 \:=\: 4 - 5x$

We have: . $6x^2 + 5x - 4 \:=\:0$

Factor: . $(2x-1)(3x+4) \:=\:0$

Therefore: . $x \:=\:\tfrac{1}{2},\;\text{-}\tfrac{4}{3}$

$[2]\;4x^2+ 2x \:=\:5$

We have: . $4x^2 + 2x - 5 \:=\:0$

Quadratic Formula: . $x \;=\;\frac{\text{-}2 \pm \sqrt{2^2 - 4(4)(\text{-}5)}}{2(4)}$

. . . . . . $x \;=\;\frac{\text{-}2\pm\sqrt{84}}{8} \;=\;\frac{\text{-}2\pm2\sqrt{21}}{8} \;=\;\frac{\text{-}1\pm\sqrt{21}}{4}$

$[3]\;9(x - 4)^2 + 1 \:=\: 0$

No fancy tricks . . . just solve for $x.$

We have: . $9(x-4)^2 \:=\:\text{-}1$

. . . . . . . . . $(x-4)^2 \:=\:\text{-}\tfrac{1}{9}{$

n . . . . . . . . . $x-4 \;=\;\pm\sqrt{\text{-}\tfrac{1}{9}}$

n . . . . . . . . . $x-4 \;=\;\pm\tfrac{1}{3}i$

n . . . . . . . . . . . . $x \;=\;4\pm\tfrac{1}{3}i$