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Question Number 187951 by chhuangakawlni last updated on 24/Feb/23

please help me to solve  6x^2 +x−2=0

$${please}\:{help}\:{me}\:{to}\:{solve} \\ $$$$\mathrm{6}{x}^{\mathrm{2}} +{x}−\mathrm{2}=\mathrm{0} \\ $$

Answered by cortano12 last updated on 24/Feb/23

(x+(4/6))(x−(3/6))=0    { ((x=−(2/3))),((x=(1/2))) :}

$$\left(\mathrm{x}+\frac{\mathrm{4}}{\mathrm{6}}\right)\left(\mathrm{x}−\frac{\mathrm{3}}{\mathrm{6}}\right)=\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{x}=−\frac{\mathrm{2}}{\mathrm{3}}}\\{\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$

Answered by Rasheed.Sindhi last updated on 24/Feb/23

6x^2 +4x−3x−2=0  2x(3x+2)−(3x+2)=0  (3x+2)(2x−1)=0  3x+2=0 ∣  2x−1=0  x=−(2/3)  ∣ x=(1/2)

$$\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{3}{x}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}{x}\left(\mathrm{3}{x}+\mathrm{2}\right)−\left(\mathrm{3}{x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\left(\mathrm{3}{x}+\mathrm{2}\right)\left(\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{3}{x}+\mathrm{2}=\mathrm{0}\:\mid\:\:\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{2}}{\mathrm{3}}\:\:\mid\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by HeferH last updated on 25/Feb/23

6x^2 + x − 2 = 0  6x^2  + 4x − 3x −2=0   2x(3x + 2) −1(3x + 2) = 0  (2x − 1)(3x + 2)=0    { ((2x−1 = 0; x = (1/2))),((3x + 2 = 0; x = −(2/3) )) :}

$$\mathrm{6}{x}^{\mathrm{2}} +\:{x}\:−\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\mathrm{6}{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:−\:\mathrm{3}{x}\:−\mathrm{2}=\mathrm{0} \\ $$$$\:\mathrm{2}{x}\left(\mathrm{3}{x}\:+\:\mathrm{2}\right)\:−\mathrm{1}\left(\mathrm{3}{x}\:+\:\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}{x}\:−\:\mathrm{1}\right)\left(\mathrm{3}{x}\:+\:\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{2}{x}−\mathrm{1}\:=\:\mathrm{0};\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{3}{x}\:+\:\mathrm{2}\:=\:\mathrm{0};\:{x}\:=\:−\frac{\mathrm{2}}{\mathrm{3}}\:}\end{cases} \\ $$

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