3x(-12x)+10=0

Solution for 3x(-12x)+10=0 equation:

3x(-12x)+10=0

We multiply parentheses

-36x^2+10=0

a = -36; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-36)·10
Δ = 1440
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{10}}{2*-36}=\frac{0-12\sqrt{10}}{-72}
=-\frac{12\sqrt{10}}{-72}
=-\frac{\sqrt{10}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{10}}{2*-36}=\frac{0+12\sqrt{10}}{-72}
=\frac{12\sqrt{10}}{-72}
=\frac{\sqrt{10}}{-6}
$