Let x

_{1} and x

_{2} = roots of the equation Ax

^{2} + Bx + C = 0

$Ax^2 + Bx + C = 0$

$x^2 + \dfrac{B}{A}x + \dfrac{C}{A} = 0$

$x^2 - \left( -\dfrac{B}{A} \right) x + \dfrac{C}{A} = 0$

From Sum and Product of Roots

$x^2 - (x_1 + x_2)x + x_1 x_2 = 0$

Let r_{1} and r_{2} = roots of the required equation

$r_1 = \dfrac{1}{x_1}$ and $r_2 = \dfrac{1}{x_2}$

$r_1 + r_2 = \dfrac{1}{x_1} + \dfrac{1}{x_2}$

$r_1 + r_2 = \dfrac{x_2 + x_1}{x_1 x_2}$

$r_1 + r_2 = \dfrac{-B/A}{C/A}$

$r_1 + r_2 = -\dfrac{B}{C}$

$r_1 r_2 = \dfrac{1}{x_1} \cdot \dfrac{1}{x_2}$

$r_1 r_2 = \dfrac{1}{x_1 x_2}$

$r_1 r_2 = \dfrac{1}{C/A}$

$r_1 r_2 = \dfrac{A}{C}$

The required equation is:

$x^2 - (r_1 + r_2)x + r_1 r_2 = 0$

$x^2 - \left( -\dfrac{B}{C} \right)x + \dfrac{A}{C} = 0$

$x^2 + \dfrac{B}{C}x + \dfrac{A}{C} = 0$

$Cx^2 + Bx + A = 0$ ^{†}

From the given equation: A = 3, B = -13, and C = -10

$-10x^2 - 13x + 3 = 0$

$10x^2 + 13x - 3 = 0$ *answer*

^{†} The quadratic equation whose roots are reciprocals to the roots of Ax^{2} + Bx + C = 0 is Cx^{2} + Bx + A = 0.