0=16x^2-22x+7

Solution for 0=16x^2-22x+7 equation:

0=16x^2-22x+7

We move all terms to the left:

0-(16x^2-22x+7)=0

We add all the numbers together, and all the variables

-(16x^2-22x+7)=0

We get rid of parentheses

-16x^2+22x-7=0

a = -16; b = 22; c = -7;
Δ = b2-4ac
Δ = 222-4·(-16)·(-7)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-6}{2*-16}=\frac{-28}{-32}
=7/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+6}{2*-16}=\frac{-16}{-32}
=1/2
$