14x^2+14x-5=0

Solution for 14x^2+14x-5=0 equation:

14x^2+14x-5=0

a = 14; b = 14; c = -5;
Δ = b2-4ac
Δ = 142-4·14·(-5)
Δ = 476
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{476}=\sqrt{4*119}=\sqrt{4}*\sqrt{119}=2\sqrt{119}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{119}}{2*14}=\frac{-14-2\sqrt{119}}{28}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{119}}{2*14}=\frac{-14+2\sqrt{119}}{28}
$