2x^2-28x+4=0

Solution for 2x^2-28x+4=0 equation:

2x^2-28x+4=0

a = 2; b = -28; c = +4;
Δ = b2-4ac
Δ = -282-4·2·4
Δ = 752
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:

$\sqrt{\Delta}=\sqrt{752}=\sqrt{16*47}=\sqrt{16}*\sqrt{47}=4\sqrt{47}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{47}}{2*2}=\frac{28-4\sqrt{47}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{47}}{2*2}=\frac{28+4\sqrt{47}}{4}
$