Question

Find the roots of the equation 3x^2-7x-6=0

Answer 1

$\Large\bf\green{Answer:}$

$\tiny{ }$

$\large\bf{To~find:}$

$\bf The \: roots \: of \: the \: equation : \\ \bf {3x}^{2} - 7x - 6 = 0$

$\bf Here : \: \: \: \: s = - 7 \: \: \: \: p = - 18 \\ \bf ( - 9 \: \: \: 2)$

$\bf{:→ (3x^{2}-9x)+(2x-6)=0}$

$\bf{:→ [3x(x-3)~]+[2(x-3)~]=0}$

$\bf{:→ (3x+2)(x-3)=0}$

$\bf{:→ 3x+2=0~~~~~or~~~~~x-3=0}$

$\bf{:→ x={\dfrac{-2}{3}}~~~~~or~~~~~x=3}$

$\bf{ \therefore The~roots~are:}$

$\bf{ {\green{{\dfrac{-2}{3}}}}~~and~~{\green{3}}}$

Answer 2

Answer:

Aɴsᴡᴇʀ:-

$\begin{gathered} \bf The \: roots \: of \: the \: equation : \\ \bf {3x}^{2} - 7x - 6 = 0\end{gathered}$

$\begin{gathered} \bf Here : \: \: \: \: s = - 7 \: \: \: \: p = - 18 \\ \bf ( - 9 \: \: \: 2) \end{gathered}$

$\bf{:→ (3x^{2}-9x)+(2x-6)=0}$

$\bf{:→ [3x(x-3)~]+[2(x-3)~]=0}:→[3x(x−3) ]+[2(x−3) ]=0$

$\bf{:→ (3x+2)(x-3)=0}:→(3x+2)(x−3)=0$

$\bf{:→ 3x+2=0~~~~~or~~~~~x-3=0}$

$\bf{:→ x={\dfrac{-2}{3}}~~~~~or~~~~~x=3}$

$\bf{ \therefore The~roots~are:}$

$\bf{ {\green{{\dfrac{-2}{3}}}}~~and~~{\green{3}}}$