2s^2+4s-10=0

Solution for 2s^2+4s-10=0 equation:

2s^2+4s-10=0

a = 2; b = 4; c = -10;
Δ = b2-4ac
Δ = 42-4·2·(-10)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{6}}{2*2}=\frac{-4-4\sqrt{6}}{4}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{6}}{2*2}=\frac{-4+4\sqrt{6}}{4}
$