Answer

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**Hint:**This question involves the operation of addition/ subtraction/ multiplication/ division. Also, we need to know the basic form of a quadratic equation and the formula to find the value of \[x\] in a quadratic equation. We need to know the square root value of basic numbers. We have \[{x^2}\] terms in the given equation so, we would find two answers \[x\] by solving the given equation.

**Complete step by step solution:**

The given question is shown below,

\[3{x^2} - 2x - 5 = 0 \to \left( 1 \right)\]

We know that the basic form of a quadratic equation is,

\[a{x^2} + bx + c = 0 \to \left( 2 \right)\]

The formula for finding the value \[x\] from the above equation is shown below,

\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 3 \right)\]

By comparing the equation \[\left( 1 \right)\]and\[\left( 2 \right)\], we get the value

of \[a,b\]and\[c\] as follows,

\[\left( 1 \right) \to 3{x^2} - 2x - 5 = 0\]

\[\left( 2 \right) \to a{x^2} + bx + c = 0\]

So, we get the value of \[a\] is \[3\], the value of \[b\]is\[ - 2\], and the value of \[c\]is\[ - 5\]. Let’s substitute these values in the equation \[\left( 3 \right)\], we get

\[\left( 3 \right) \to x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

\[x = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4 \times 3 \times - 5} }}{{2 \times 3}}\]

So, we get

\[

x = \dfrac{{2 \pm \sqrt {4 + 60} }}{6} \\

x = \dfrac{{2 \pm \sqrt {64} }}{6} \\

\]

We know that, \[{8^2} = 64\]. So, the above equation can also be written as,

\[

x = \dfrac{{2 \pm \sqrt {{8^2}} }}{6} \\

x = \dfrac{{2 \pm 8}}{6} \\

\]

So, the above equation can be solved by using two cases.

Case: \[1\]

\[

x = \dfrac{{2 + 8}}{6} \\

x = \dfrac{{10}}{6} \\

x = \dfrac{5}{3} \\

\]

Case: \[2\]

\[

x = \dfrac{{2 - 8}}{6} \\

x = \dfrac{{ - 6}}{6} \\

x = - 1 \\

\]

**So, the final answer is**

\[x = - 1\]and\[x = \dfrac{5}{3}\].

\[x = - 1\]and\[x = \dfrac{5}{3}\].

**Note:**This type of question involves the operation of addition/ subtraction/ multiplication/ division. Note that the denominator would not be equal to zero. When \[{n^2}\] is placed inside the square root we can cancel the square root and square with each other. Note that if \[ \pm \]is present in the equation we would find two values for \[x\]. Also, remember the basic form of a quadratic equation

and the formula for finding the value \[x\].

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