14n^2+2n-20=0

Solution for 14n^2+2n-20=0 equation:

14n^2+2n-20=0

a = 14; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·14·(-20)
Δ = 1124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1124}=\sqrt{4*281}=\sqrt{4}*\sqrt{281}=2\sqrt{281}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{281}}{2*14}=\frac{-2-2\sqrt{281}}{28}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{281}}{2*14}=\frac{-2+2\sqrt{281}}{28}
$