0=-16x^2-18x

Solution for 0=-16x^2-18x equation:

0=-16x^2-18x

We move all terms to the left:

0-(-16x^2-18x)=0

We add all the numbers together, and all the variables

-(-16x^2-18x)=0

We get rid of parentheses

16x^2+18x=0

a = 16; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·16·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*16}=\frac{-36}{32}
=-1+1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*16}=\frac{0}{32}
=0
$