How do you solve \[3{x^2} + 7x - 13 = 0\] ?

Hint: A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. In the given question we have to solve the given quadratic equation using the quadratic formula. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .

Complete step-by-step answer:
Given, \[3{x^2} + 7x - 13 = 0\] .
We know the standard quadratic equation is of the form \[a{x^2} + bx + c = 0\] . Comparing the given problem we have \[a = 3\] , \[b = 7\] and \[c = - 13\] .
We can see that we cannot expand the middle term into a sum of two terms. Because the product \[ac = - 39\] and the factors of 39 are 13 and 3 only. We cannot find two terms such that their product is \[ - 39\] and sum is 7.
So we use the quadratic formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .
Substituting the values we have,
 \[x = \dfrac{{ - 7 \pm \sqrt {{7^2} - 4 \times (3) \times ( - 13)} }}{{2(3)}}\]
We know that negative number multiplied by negative number we will have a positive number,
 \[ = \dfrac{{ - 7 \pm \sqrt {49 + 156} }}{6}\]
 \[ = \dfrac{{ - 7 \pm \sqrt {205} }}{6}\] , we cannot simplify further. Because 205 is not divided by a perfect square number.
Hence we have,
 \[ \Rightarrow x = \dfrac{{ - 7 + \sqrt {205} }}{6}\] and \[x = \dfrac{{ - 7 - \sqrt {205} }}{6}\] .
So, the correct answer is “ \[ x = \dfrac{{ - 7 + \sqrt {205} }}{6}\] and \[x = \dfrac{{ - 7 - \sqrt {205} }}{6}\] ”.

Note: Since the degree of the given polynomial is 2. Hence we have two roots. In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. The quadratic formula is also called Sridhar’s formula. Careful in the calculation part.