s^2+2s-243=0

Solution for s^2+2s-243=0 equation:

s^2+2s-243=0

a = 1; b = 2; c = -243;
Δ = b2-4ac
Δ = 22-4·1·(-243)
Δ = 976
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{976}=\sqrt{16*61}=\sqrt{16}*\sqrt{61}=4\sqrt{61}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{61}}{2*1}=\frac{-2-4\sqrt{61}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{61}}{2*1}=\frac{-2+4\sqrt{61}}{2}
$