Question

solve the equation 3x^2-4x+20/3=0​

Answer 1

Answer:

Your answer in the above picture

Answer 2

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\sf Solve \:the \:equation:$

$\sf \bf 3{x}^{2}-4x+\dfrac{20}{3}=0$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf\implies 3{x}^{2}-4x+\dfrac{20}{3}=0$

${\underbrace {\overbrace {\color {orange} {\bf Multiply \:both \:sides \:by\:3}}}}$

$\sf \bf\implies 3\bigg( 3{x}^{2}-4x+\dfrac{20}{3}\bigg)=3\bigg(0\bigg)$

$\sf \bf\implies 3 \times 3{x}^{2}-3 \times 4x+\cancel {3} \times \dfrac{20}{\cancel {3}}=3 \times 0$

$\sf \bf\implies 9{x}^{2}-12x+20=0$

${\color{gold} {\underline{ \sf It \: is \: in \: the \: form \: of \: {{\underbrace \color{red} \: a{x}^{2} +bx+c=0}}}}}$

${\boxed {\boxed {\color {blue} { \sf \bf x=\dfrac{-b\pm\sqrt{{b}^{2}-4ac}}{2a}}}}}$

$\sf \bf a=9\\\sf \bf b=\Big(-12\Big)\\\sf \bf c=20$

${\underbrace {\overbrace {\color {green} {\sf\bf Substitute \:the \:values}}}}$

$\sf \bf\implies x=\dfrac{-\Big(-12\Big)\pm\sqrt{{\Big(-12\Big)}^{2}-4\Big(9\Big)\Big(20\Big)}}{2\Big(9\Big)}$

$\sf \bf\implies x=\dfrac{12\pm\sqrt{144-720}}{18}$

$\sf \bf\implies x=\dfrac{12\pm\sqrt{-576}}{18}$

$\sf \bf\implies x=\dfrac{12\pm\sqrt{-1 \times 576}}{18}$

$\sf \bf\implies x=\dfrac{12\pm\sqrt{-1}\times \sqrt{576}}{18}$

$\sf \bf\implies x=\dfrac{12\pm i\times 24}{18}$

$\sf \bf\implies x=\dfrac{12\pm 24i}{18}$

$\sf \bf\implies x=\dfrac{6\Big(2\pm 4i\Big)}{18}$

$\sf \bf\implies x=\dfrac{2\pm 4i}{3}$

$\implies{\boxed {\color {purple} { \sf \bf x=\dfrac{2+4i}{3}\:\:or\:\:x=\dfrac{2-4i}{3}}}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

${\color {red}{\tt Quadratic \:equation}}$

• $\sf Second\: degree\: equations\: are \:called\\\sf quadratic\: equations.$
• $\sf a{x}^{2}+bx+c=0\: is\: the\: standard\: form\: of \\\sf quadratic\: equation\: of\: variable\: x. a, b, c\\\sf are\: constants\: and\: a\neq 0$