3=-16x^2+45x

Solution for 3=-16x^2+45x equation:

3=-16x^2+45x

We move all terms to the left:

3-(-16x^2+45x)=0

We get rid of parentheses

16x^2-45x+3=0

a = 16; b = -45; c = +3;
Δ = b2-4ac
Δ = -452-4·16·3
Δ = 1833
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{1833}}{2*16}=\frac{45-\sqrt{1833}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{1833}}{2*16}=\frac{45+\sqrt{1833}}{32}
$