0=-16x^2+92x

Solution for 0=-16x^2+92x equation:

0=-16x^2+92x

We move all terms to the left:

0-(-16x^2+92x)=0

We add all the numbers together, and all the variables

-(-16x^2+92x)=0

We get rid of parentheses

16x^2-92x=0

a = 16; b = -92; c = 0;
Δ = b2-4ac
Δ = -922-4·16·0
Δ = 8464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8464}=92$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-92)-92}{2*16}=\frac{0}{32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-92)+92}{2*16}=\frac{184}{32}
=5+3/4
$