$$3x^2 + 4x - 2 = 0 \qquad x?$$
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: $$x_1+x_2= 0.38742588672-1.72075922006=-1.\bar{3}=-p$$
$$x_1x_2=0.38742588672\times(-1.72075922006)=-0.\bar{6}=q$$
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