2q2+14q-5=0

Solution for 2q2+14q-5=0 equation:

2q^2+14q-5=0

a = 2; b = 14; c = -5;
Δ = b2-4ac
Δ = 142-4·2·(-5)
Δ = 236
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{236}=\sqrt{4*59}=\sqrt{4}*\sqrt{59}=2\sqrt{59}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{59}}{2*2}=\frac{-14-2\sqrt{59}}{4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{59}}{2*2}=\frac{-14+2\sqrt{59}}{4}
$