s^2=35=-35

Solution for s^2=35=-35 equation:

s^2=35=-35

We move all terms to the left:

s^2-(35)=0

a = 1; b = 0; c = -35;
Δ = b2-4ac
Δ = 02-4·1·(-35)
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{35}}{2*1}=\frac{0-2\sqrt{35}}{2}
=-\frac{2\sqrt{35}}{2}
=-\sqrt{35}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{35}}{2*1}=\frac{0+2\sqrt{35}}{2}
=\frac{2\sqrt{35}}{2}
=\sqrt{35}
$