16x^2-22x+6=0

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Solution for 16x^2-22x+6=0 equation: 



16x^2-22x+6=0

a = 16; b = -22; c = +6;
Δ = b2-4ac
Δ = -222-4·16·6
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-10}{2*16}=\frac{12}{32}
=3/8
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+10}{2*16}=\frac{32}{32}
=1
$

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