19x^2-36x+4=0

Solution for 19x^2-36x+4=0 equation:

19x^2-36x+4=0

a = 19; b = -36; c = +4;
Δ = b2-4ac
Δ = -362-4·19·4
Δ = 992
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:

$\sqrt{\Delta}=\sqrt{992}=\sqrt{16*62}=\sqrt{16}*\sqrt{62}=4\sqrt{62}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{62}}{2*19}=\frac{36-4\sqrt{62}}{38}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{62}}{2*19}=\frac{36+4\sqrt{62}}{38}
$