130+48x-3x^2=0

Solution for 130+48x-3x^2=0 equation:

130+48x-3x^2=0

a = -3; b = 48; c = +130;
Δ = b2-4ac
Δ = 482-4·(-3)·130
Δ = 3864
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:

$\sqrt{\Delta}=\sqrt{3864}=\sqrt{4*966}=\sqrt{4}*\sqrt{966}=2\sqrt{966}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-2\sqrt{966}}{2*-3}=\frac{-48-2\sqrt{966}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+2\sqrt{966}}{2*-3}=\frac{-48+2\sqrt{966}}{-6}
$