Question

Solve the quadratic equation $10x^2+x-3=0$ by using the quadratic formula.

Answer 1

Determining the value of $x$ by using the quadratic formula:

The quadratic formula is $\begin{array}{rcl}x& =& \frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$.

For the given equation $10x^2+x-3=0$, the value of variables are $a=10,b=1$ and $c=-3$.

Substituting the values of $a,b$ and $c$ in quadratic formula,

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \Rightarrow x=\frac{-1\pm \sqrt{{\left(1\right)}^2-4\left(10\right)(-3)}}{2\left(10\right)} \\ \Rightarrow x=\frac{-1\pm \sqrt{1+120}}{20} \\ \Rightarrow x=\frac{-1\pm \sqrt{121}}{20} \\ \Rightarrow x=\frac{-1\pm 11}{20} \\ \Rightarrow x=\frac{-1-11}{20},\frac{-1+11}{20} \\ \Rightarrow x=-\frac{12}{20},\frac{10}{20} \end{gathered}$$

$\therefore x=-\frac{3}{5},\frac{1}{2}$

Therefore, the roots of the given equation are $-\frac{3}{5}$, $\frac{1}{2}$.