19w^2=19w+14

Solution for 19w^2=19w+14 equation:

19w^2=19w+14

We move all terms to the left:

19w^2-(19w+14)=0

We get rid of parentheses

19w^2-19w-14=0

a = 19; b = -19; c = -14;
Δ = b2-4ac
Δ = -192-4·19·(-14)
Δ = 1425
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1425}=\sqrt{25*57}=\sqrt{25}*\sqrt{57}=5\sqrt{57}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5\sqrt{57}}{2*19}=\frac{19-5\sqrt{57}}{38}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5\sqrt{57}}{2*19}=\frac{19+5\sqrt{57}}{38}
$