14u2+23u=3

Solution for 14u2+23u=3 equation:

14u^2+23u=3

We move all terms to the left:

14u^2+23u-(3)=0

a = 14; b = 23; c = -3;
Δ = b2-4ac
Δ = 232-4·14·(-3)
Δ = 697
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{697}}{2*14}=\frac{-23-\sqrt{697}}{28}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{697}}{2*14}=\frac{-23+\sqrt{697}}{28}
$