40=-16x^2+96x+3

Solution for 40=-16x^2+96x+3 equation:

40=-16x^2+96x+3

We move all terms to the left:

40-(-16x^2+96x+3)=0

We get rid of parentheses

16x^2-96x-3+40=0

We add all the numbers together, and all the variables

16x^2-96x+37=0

a = 16; b = -96; c = +37;
Δ = b2-4ac
Δ = -962-4·16·37
Δ = 6848
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{6848}=\sqrt{64*107}=\sqrt{64}*\sqrt{107}=8\sqrt{107}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-8\sqrt{107}}{2*16}=\frac{96-8\sqrt{107}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+8\sqrt{107}}{2*16}=\frac{96+8\sqrt{107}}{32}
$