x=-16x^2+18x

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

x=-16x^2+18x

We move all terms to the left:

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

We get rid of parentheses

16x^2-18x+x=0

We add all the numbers together, and all the variables

16x^2-17x=0

a = 16; b = -17; c = 0;
Δ = b2-4ac
Δ = -172-4·16·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-17}{2*16}=\frac{0}{32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+17}{2*16}=\frac{34}{32}
=1+1/16
$