12x^2+48x-3=0

Solution for 12x^2+48x-3=0 equation:

12x^2+48x-3=0

a = 12; b = 48; c = -3;
Δ = b2-4ac
Δ = 482-4·12·(-3)
Δ = 2448
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:

$\sqrt{\Delta}=\sqrt{2448}=\sqrt{144*17}=\sqrt{144}*\sqrt{17}=12\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-12\sqrt{17}}{2*12}=\frac{-48-12\sqrt{17}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+12\sqrt{17}}{2*12}=\frac{-48+12\sqrt{17}}{24}
$