I hope the choices for the numerators of the solutions are given.
I am showing the complete work to find the solutions of this equation , it will help you to find an answer of your question based on this solution.
The standard form of a quadratic equation is :
ax² + bx + c = 0
And the quadratic formula is:
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
So, first step is to compare the given equation with the above equation to get the value of a, b and c.
So, a = 10, b = -19 and c = 6.
Next step is to plug in these values in the above formula. Therefore,
\(x=\frac{(-19)-\pm\sqrt{(-19)^2-4*10*6}}{2*10}\)
\(=\frac{19\pm\sqrt{361-240}}{20}\)
\(=\frac{19\pm\sqrt{121}}{20}\)
\(=\frac{19\pm11}{20}\)
So, \(x=\frac{19-11}{20} ,\frac{19+11}{20}\)
\(x=\frac{8}{20} , \frac{30}{20}\)
So, \(x= \frac{2}{5} ,\frac{3}{2}\)
Hope this helps you!
