How do you solve $3{x^2} + 5x = 0$ using the quadratic formula?

Hint: First compare the given quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.

Formula used:
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step answer:
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
First, compare $3{x^2} + 5x = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing $3{x^2} + 5x = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 3$, $b = 5$ and $c = 0$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( 5 \right)^2} - 4\left( 3 \right)\left( 0 \right)$
After simplifying the result, we get
$ \Rightarrow D = 25$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$x = \dfrac{{ - 5 \pm 5}}{{2 \times 3}}$
It can be written as
$ \Rightarrow x = \dfrac{{ - 5 \pm 5}}{6}$
$ \Rightarrow x = 0$ and $x = \dfrac{{ - 10}}{6} = \dfrac{{ - 5}}{3}$
So, $x = 0$ and $x = - \dfrac{5}{3}$ are roots/solutions of equation $3{x^2} + 5x = 0$.

Therefore, the solutions to the quadratic equation $3{x^2} + 5x = 0$ are $x = 0$ and $x = - \dfrac{5}{3}$.

Note: We can check whether $x = 0$ and $x = - \dfrac{5}{3}$ are roots/solutions of equation $2{x^2} - 5x = - 7$ by putting the value of $x$ in given equation.
Putting $x = 0$ in LHS of equation $3{x^2} + 5x = 0$.
\[{\text{LHS}} = 3{\left( 0 \right)^2} + 5\left( 0 \right)\]
On simplification, we get
\[ \Rightarrow {\text{LHS}} = 0\]
$\therefore {\text{LHS}} = {\text{RHS}}$
Thus, $x = 0$ is a solution of equation $3{x^2} + 5x = 0$.
Putting $x = - \dfrac{5}{3}$ in LHS of equation $3{x^2} + 5x = 0$.
\[{\text{LHS}} = 3{\left( { - \dfrac{5}{3}} \right)^2} + 5\left( { - \dfrac{5}{3}} \right)\]
On simplification, we get
\[ \Rightarrow {\text{LHS}} = \dfrac{{25}}{3} - \dfrac{{25}}{3}\]
\[ \Rightarrow {\text{LHS}} = 0\]
$\therefore {\text{LHS}} = {\text{RHS}}$
Thus, $x = - \dfrac{5}{3}$ is a solution of equation $3{x^2} + 5x = 0$.
Final solution: Therefore, the solutions to the quadratic equation $3{x^2} + 5x = 0$ are $x = 0$ and $x = - \dfrac{5}{3}$.