l^2+14l-126=0

Solution for l^2+14l-126=0 equation:

l^2+14l-126=0

a = 1; b = 14; c = -126;
Δ = b2-4ac
Δ = 142-4·1·(-126)
Δ = 700
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-10\sqrt{7}}{2*1}=\frac{-14-10\sqrt{7}}{2}
$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+10\sqrt{7}}{2*1}=\frac{-14+10\sqrt{7}}{2}
$