19x^2-38x+7=0

Solution for 19x^2-38x+7=0 equation:

19x^2-38x+7=0

a = 19; b = -38; c = +7;
Δ = b2-4ac
Δ = -382-4·19·7
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-4\sqrt{57}}{2*19}=\frac{38-4\sqrt{57}}{38}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+4\sqrt{57}}{2*19}=\frac{38+4\sqrt{57}}{38}
$