4x^2-22x+14=0

Solution for 4x^2-22x+14=0 equation:

4x^2-22x+14=0

a = 4; b = -22; c = +14;
Δ = b2-4ac
Δ = -222-4·4·14
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{65}}{2*4}=\frac{22-2\sqrt{65}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{65}}{2*4}=\frac{22+2\sqrt{65}}{8}
$