How do you solve $3{{x}^{2}}+13x-10=0?$

Hint: The given equation is called a polynomial equation with second-degree and is also known as the quadratic equation and the standard form of the second-degree polynomial equation is $a{{x}^{2}}+bx+c=0.$where $a$and $b$are the coefficients of the ${{x}^{2}}$and $x$, and $c$is the constant value.
The polynomial with equality is called equations for example$a{{x}^{2}}+4x=9$.
To solve this type of equation we will use the quadratic formula state that the value of $x$that’s
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
In the above formula $b$is the coefficient of $x$ and $a$is coefficient of ${{x}^{2}}$, $c$is constant.

Complete step by step solution:
The given quadratics equation is $3{{x}^{2}}+13x-10=0$
After comparing the above equation with the standard form of the quadratic equation that’s $a{{x}^{2}}+bx+c=0$ we get that $a=3$,$b=13$and $c=-10$
Let’s put the value of all coefficients in the quadratic formula
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$\Rightarrow x=\dfrac{-13\pm \sqrt{{{13}^{2}}-4\times 3\times \left( -10 \right)}}{2\times 3}$
\[\Rightarrow x=\dfrac{-13\pm \sqrt{169-(-120)}}{6}\]
\[\Rightarrow x=\dfrac{-13\pm \sqrt{169+120}}{6}\]
\[\Rightarrow x=\dfrac{-13\pm \sqrt{289}}{6}\]
\[\Rightarrow x=\dfrac{-13\pm 17}{6}\]
We rewrite the above term in two-part as $x=\dfrac{-13+17}{6}$ and $x=\dfrac{-13-17}{6}$
If $x=\dfrac{-13+17}{6}$then $x$will be $\dfrac{4}{6}$that can be written as $\dfrac{2}{3}$after simplifying this fraction by 2
And if $x=\dfrac{-13-17}{6}$the x will be $\dfrac{-30}{6}$that can be written as $-5$.

Hence, the values of $x$will be $\dfrac{2}{3},-5$.

Note:
1) The shape of the quadratic polynomial equation is always parabolic.
2) The roots of the quadratic polynomials equation are two real, zero, or one.
3) If the degree is the highest power of the variables of the polynomial equation then it’s called a quadratic polynomial equation.
4) If the degree is one then the polynomial equation will be linear.