35x^2-68x+19=0

Solution for 35x^2-68x+19=0 equation:

35x^2-68x+19=0

a = 35; b = -68; c = +19;
Δ = b2-4ac
Δ = -682-4·35·19
Δ = 1964
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:

$\sqrt{\Delta}=\sqrt{1964}=\sqrt{4*491}=\sqrt{4}*\sqrt{491}=2\sqrt{491}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-68)-2\sqrt{491}}{2*35}=\frac{68-2\sqrt{491}}{70}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-68)+2\sqrt{491}}{2*35}=\frac{68+2\sqrt{491}}{70}
$