2x^2+6x-243=0

Solution for 2x^2+6x-243=0 equation:

2x^2+6x-243=0

a = 2; b = 6; c = -243;
Δ = b2-4ac
Δ = 62-4·2·(-243)
Δ = 1980
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:

$\sqrt{\Delta}=\sqrt{1980}=\sqrt{36*55}=\sqrt{36}*\sqrt{55}=6\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{55}}{2*2}=\frac{-6-6\sqrt{55}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{55}}{2*2}=\frac{-6+6\sqrt{55}}{4}
$