14q^2+12q-23=0

Solution for 14q^2+12q-23=0 equation:

14q^2+12q-23=0

a = 14; b = 12; c = -23;
Δ = b2-4ac
Δ = 122-4·14·(-23)
Δ = 1432
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:

$\sqrt{\Delta}=\sqrt{1432}=\sqrt{4*358}=\sqrt{4}*\sqrt{358}=2\sqrt{358}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{358}}{2*14}=\frac{-12-2\sqrt{358}}{28}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{358}}{2*14}=\frac{-12+2\sqrt{358}}{28}
$