s^2+12s-35=0

Solution for s^2+12s-35=0 equation:

s^2+12s-35=0

a = 1; b = 12; c = -35;
Δ = b2-4ac
Δ = 122-4·1·(-35)
Δ = 284
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:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{71}}{2*1}=\frac{-12-2\sqrt{71}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{71}}{2*1}=\frac{-12+2\sqrt{71}}{2}
$