3x^2+6x-1053=0

Solution for 3x^2+6x-1053=0 equation:

3x^2+6x-1053=0

a = 3; b = 6; c = -1053;
Δ = b2-4ac
Δ = 62-4·3·(-1053)
Δ = 12672
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:

$\sqrt{\Delta}=\sqrt{12672}=\sqrt{576*22}=\sqrt{576}*\sqrt{22}=24\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-24\sqrt{22}}{2*3}=\frac{-6-24\sqrt{22}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+24\sqrt{22}}{2*3}=\frac{-6+24\sqrt{22}}{6}
$