0=-16x^2+45x

Solution for 0=-16x^2+45x equation:

0=-16x^2+45x

We move all terms to the left:

0-(-16x^2+45x)=0

We add all the numbers together, and all the variables

-(-16x^2+45x)=0

We get rid of parentheses

16x^2-45x=0

a = 16; b = -45; c = 0;
Δ = b2-4ac
Δ = -452-4·16·0
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-45}{2*16}=\frac{0}{32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+45}{2*16}=\frac{90}{32}
=2+13/16
$