0=16x^2-22x-3

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

0=16x^2-22x-3

We move all terms to the left:

0-(16x^2-22x-3)=0

We add all the numbers together, and all the variables

-(16x^2-22x-3)=0

We get rid of parentheses

-16x^2+22x+3=0

a = -16; b = 22; c = +3;
Δ = b2-4ac
Δ = 222-4·(-16)·3
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-26}{2*-16}=\frac{-48}{-32}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+26}{2*-16}=\frac{4}{-32}
=-1/8
$