(-3x^2)+2x+48=0

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

(-3x^2)+2x+48=0

We get rid of parentheses

-3x^2+2x+48=0

a = -3; b = 2; c = +48;
Δ = b2-4ac
Δ = 22-4·(-3)·48
Δ = 580
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:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{145}}{2*-3}=\frac{-2-2\sqrt{145}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{145}}{2*-3}=\frac{-2+2\sqrt{145}}{-6}
$