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How do you solve $3{{x}^{2}}+7x-6=0$?

Answer
VerifiedVerified
543.3k+ views
Hint: For solving the equation given in the above question, which is a quadratic equation, we need to use the quadratic formula which is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. For this, we need to substitute the values of the coefficients a, b, and c from the given equation $3{{x}^{2}}+7x-6=0$. In the equation $3{{x}^{2}}+7x-6=0$, we can observe that $a=3$, $b=7$ and $c=-6$. On substituting these values of the coefficients into the quadratic formula, we will obtain the two solutions of the given equation.

Complete step by step answer:
The equation given in the above question is written as
$\Rightarrow 3{{x}^{2}}+7x-6=0$
Since the highest power of the variable x in the above equation is equal to two, it is a quadratic equation. Therefore we can use the quadratic formula for solving it, which is givne by
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
From the given equation, we can observe the below value of the coefficients.
$\begin{align}
  & \Rightarrow a=3 \\
 & \Rightarrow b=7 \\
 & \Rightarrow c=-6 \\
\end{align}$
Substituting these values into the quadratic formula, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-7\pm \sqrt{{{\left( 7 \right)}^{2}}-4\left( 3 \right)\left( -6 \right)}}{2\left( 3 \right)} \\
 & \Rightarrow x=\dfrac{-7\pm \sqrt{49+72}}{6} \\
 & \Rightarrow x=\dfrac{-7\pm \sqrt{121}}{6} \\
 & \Rightarrow x=\dfrac{-7\pm 11}{6} \\
 & \Rightarrow x=\dfrac{-7+11}{6},x=\dfrac{-7-11}{6} \\
 & \Rightarrow x=\dfrac{4}{6},x=\dfrac{-18}{6} \\
 & \Rightarrow x=\dfrac{2}{3},x=-3 \\
\end{align}\]

Hence, we have obtained the two solutions of the given quadratic equation as \[x=\dfrac{2}{3}\] and \[x=-3\].

Note: We can also use the other methods for solving the quadratic equation given in the question. For example, we can use the middle term splitting method, or the completing the square method. Take proper care of the signs of the coefficients before substituting them into the quadratic formula. The value of c is negative, and therefore, we must not ignore it.