How do you solve $3{x^2} + 2x - 5 = 0$

Hint:In order to solve this question there are two methods which we apply here for solving it first one is by factoring all the equations and then we will put it equal to 0 and from it we will get the values of x. There is another approach in which we will directly apply the dharacharya formula and by putting the values of coefficients we will get the final answer.

Step by step solution:
For solving this we will have to factorize the equation:
The equation is:
$3{x^2} + 2x - 5 = 0$
On breaking 2x into parts:
$3{x^2} + 5x - 3x - 5 = 0$
Taking 3x common in 1st and 3rd terms and 5 in 2nd and 4th terms:
$3x(x - 1) + 5(x - 1) = 0$
Now we will be taking (x-1) common from both terms:
$\left( {x - 1} \right)\left( {3x + 5} \right) = 0$
Since two numbers are in multiplication and their product is 0 so either of them will be zero.
$
  x - 1 = 0 \\
  3x + 5 = 0 \\
 $
From solving these two equations we will get the value of x=1 and $x = - \dfrac{5}{3}$
Alternate method:
For this method we need to have a knowledge about the dharacharya formula for finding the roots of the quadratic equation;
So dharacharya formula is given as:
If the equation is $a{x^2} + bx + c = 0$ then the roots of this quadratic equation can be calculated by the formula:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now according to our question the value of:
a=3, b=2 and c=-5
putting these values in formula:
$x = \dfrac{{ - 2 \pm \sqrt {{2^2} - \{ 4 \times 3 \times ( - 5)\} } }}{{2 \times 3}}$
On solving this we will get two values of x which are $x = 1, - \dfrac{5}{3}$

Note: While solving these types of problems when we are using factorization methods we should keep in mind that the factors are made in such a way that it may make pair and take the common carefully. While we are using dharacharya the important point is to keep the correct value in the correct position and solve it carefully because there is root and solving it as we are new to this cause the mistake. If we want to check whether our answer is correct or not then put the value of x in the equation it will satisfy or else the founded roots are wrong.