16x^2-22x+3=0

Solution for 16x^2-22x+3=0 equation:

16x^2-22x+3=0

a = 16; b = -22; c = +3;
Δ = b2-4ac
Δ = -222-4·16·3
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{73}}{2*16}=\frac{22-2\sqrt{73}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{73}}{2*16}=\frac{22+2\sqrt{73}}{32}
$