0=16x^2-22x-27

Solution for 0=16x^2-22x-27 equation:

0=16x^2-22x-27

We move all terms to the left:

0-(16x^2-22x-27)=0

We add all the numbers together, and all the variables

-(16x^2-22x-27)=0

We get rid of parentheses

-16x^2+22x+27=0

a = -16; b = 22; c = +27;
Δ = b2-4ac
Δ = 222-4·(-16)·27
Δ = 2212
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:

$\sqrt{\Delta}=\sqrt{2212}=\sqrt{4*553}=\sqrt{4}*\sqrt{553}=2\sqrt{553}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{553}}{2*-16}=\frac{-22-2\sqrt{553}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{553}}{2*-16}=\frac{-22+2\sqrt{553}}{-32}
$