$$3x^2 + 4x - 2 = 0 \qquad x?$$
An easy way!
"p,q-formula": $$x^2+px+q=0$$
solution: $$x_{1,2}=-\dfrac{p}{2}\pm\sqrt{\dfrac{p^2}{4}-q}$$
check: $$x_1+x_2=-p\qquad x_1x_2=q$$
formula: $$3x^2 + 4x - 2 = 0$$
prepare for "p,q-formula": $$3x^2 + 4x - 2 = 0 \quad | \quad :3$$
$$x^2+\dfrac{4}{3}x-\dfrac{2}{3} =0 \qquad p= \dfrac{4}{3} \qquad q=-\dfrac{2}{3}$$
solution: $$x_{1,2}=-\dfrac{1}{2}\left(\dfrac{4}{3}\right)\pm\sqrt{\dfrac{1}{4}
\left(\dfrac{4}{3}\right)^2 +\dfrac{2}{3}}$$
$$x_{1,2}=-\dfrac{2}{3}\pm\sqrt{\dfrac{4}{3*3}+\dfrac{2*3}{3*3}}$$
$$x_{1,2}=-\dfrac{2}{3}\pm\dfrac{\sqrt{10} }{3}$$
$$x=x_1=\dfrac{-2+\sqrt{10}}{3} =0.38742588672$$
$$x=x_2=\dfrac{-2-\sqrt{10}}{3}=-1.72075922006$$
check: $$x_1+x_2= 0.38742588672-1.72075922006=-1.\bar{3}=-p$$
$$x_1x_2=0.38742588672\times(-1.72075922006)=-0.\bar{6}=q$$
.3x^2 + 4x - 2 = 0 x?
$$\\3x^2+4x-2=0\\\\
3(x^2+\frac{4}{3}x) =2 \quad | \quad :3\\\\
x^2+\frac{4}{3}x =\frac{2}{3}\\\\
\underbrace{x^2+\frac{4}{3}x+ (\frac{4}{2*3})^2}_{
=(x+\frac{4}{2*3})^2
} -(\frac{4}{2*3})^2 =\frac{2}{3}\\\\
(x+\frac{4}{2*3})^2 -(\frac{4}{2*3})^2 = \frac{2}{3}\\\\
(x+\frac{2}{3})^2 -(\frac{2}{3})^2 = \frac{2}{3}\\\\
(x+\frac{2}{3})^2 =(\frac{2}{3})^2 + \frac{2}{3}\\\\$$
$$\\(x+\frac{2}{3})^2 = \frac{2}{3}(\frac{2}{3}+1)\\\\
(x+\frac{2}{3})^2 = \frac{2}{3}*\frac{5}{3} \quad | \quad \pm\sqrt{}\\\\
x+\frac{2}{3} = \pm \sqrt{\frac{2}{3}*\frac{5}{3}}\\\\
x+\frac{2}{3} = \pm \frac{\sqrt{10}}{3}\\\\
x = -\frac{2}{3} \pm \frac{\sqrt{10}}{3}\\\\
x=x_1=\frac{-2+\sqrt{10}}{3}\\\\
x=x_2=\frac{-2-\sqrt{10}}{3}$$
$$x_1=0.38742588672$$
$$x_2=-1.72075922006$$
heureka has shown the method known as "completing the square"...we can also use the "quadratic formula"
It's given by
[-b ± √(b^2 - 4ac)] / (2a) where, in our case..... a= 3 b = 4 and c = -2 so we have....
[ -4 ± √(4^2 - 4(3)(-2))] / (2*3) =
[ -4 ± √(16 + 24)] / (6) =
[ -4 ± √(40)] / (6) =
[ -4 ± 2√(10)] / (6) (divide everything by 2) =
[ -2 ± √(10)] / (3)
Note that this is exactly the same answer(s) as heureka's !!!!
$$3x^2 + 4x - 2 = 0 \qquad x?$$
An easy way!
"p,q-formula": $$x^2+px+q=0$$
solution: $$x_{1,2}=-\dfrac{p}{2}\pm\sqrt{\dfrac{p^2}{4}-q}$$
check: $$x_1+x_2=-p\qquad x_1x_2=q$$
formula: $$3x^2 + 4x - 2 = 0$$
prepare for "p,q-formula": $$3x^2 + 4x - 2 = 0 \quad | \quad :3$$
$$x^2+\dfrac{4}{3}x-\dfrac{2}{3} =0 \qquad p= \dfrac{4}{3} \qquad q=-\dfrac{2}{3}$$
solution: $$x_{1,2}=-\dfrac{1}{2}\left(\dfrac{4}{3}\right)\pm\sqrt{\dfrac{1}{4}
\left(\dfrac{4}{3}\right)^2 +\dfrac{2}{3}}$$
$$x_{1,2}=-\dfrac{2}{3}\pm\sqrt{\dfrac{4}{3*3}+\dfrac{2*3}{3*3}}$$
$$x_{1,2}=-\dfrac{2}{3}\pm\dfrac{\sqrt{10} }{3}$$
$$x=x_1=\dfrac{-2+\sqrt{10}}{3} =0.38742588672$$
$$x=x_2=\dfrac{-2-\sqrt{10}}{3}=-1.72075922006$$
check: $$x_1+x_2= 0.38742588672-1.72075922006=-1.\bar{3}=-p$$
$$x_1x_2=0.38742588672\times(-1.72075922006)=-0.\bar{6}=q$$
The quadratic formula, that CPhill used is definitely the way to go. Why make it hardeer than necessary.
(The other answers are good value too)
This will help you remember it.
Heureka, your method is definitely interesting......whether it's "easy" could be up for debate ...!!!
But......as I always say, if it works for you....go for it !!!!