ANSWER
[tex]x=\frac{2}{5};x=-\frac{1}{3}[/tex]
EXPLANATION
We want to solve the given quadratic equation by using the quadratic formula:
[tex]15x^2-x-2=0[/tex]
The quadratic formula is:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
where a = coefficient of x²
b = coefficient of x
c = constant
From the given equation, we have that:
[tex]a=15,b=-1,c=-2[/tex]
Therefore, solving using the quadratic formula, we have:
[tex]\begin{gathered} x=\frac{-(-1)\pm\sqrt[]{(-1)^2-4(15)(-2)}}{2(15)} \\ x=\frac{1\pm\sqrt[]{1+120}}{30} \\ x=\frac{1\pm\sqrt[]{121}}{30}=\frac{1\pm11}{30} \\ \Rightarrow x=\frac{1+11}{30};x=\frac{1-11}{30} \\ \Rightarrow x=\frac{12}{30};x=\frac{-10}{30} \\ x=\frac{2}{5};x=-\frac{1}{3} \end{gathered}[/tex]
That is the solution of the quadratic equation.
