45=-16x^2+45x+40

Solution for 45=-16x^2+45x+40 equation:

45=-16x^2+45x+40

We move all terms to the left:

45-(-16x^2+45x+40)=0

We get rid of parentheses

16x^2-45x-40+45=0

We add all the numbers together, and all the variables

16x^2-45x+5=0

a = 16; b = -45; c = +5;
Δ = b2-4ac
Δ = -452-4·16·5
Δ = 1705
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-\sqrt{1705}}{2*16}=\frac{45-\sqrt{1705}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{1705}}{2*16}=\frac{45+\sqrt{1705}}{32}
$