10=-16x^2+40x+3

Solution for 10=-16x^2+40x+3 equation:

10=-16x^2+40x+3

We move all terms to the left:

10-(-16x^2+40x+3)=0

We get rid of parentheses

16x^2-40x-3+10=0

We add all the numbers together, and all the variables

16x^2-40x+7=0

a = 16; b = -40; c = +7;
Δ = b2-4ac
Δ = -402-4·16·7
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-24\sqrt{2}}{2*16}=\frac{40-24\sqrt{2}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+24\sqrt{2}}{2*16}=\frac{40+24\sqrt{2}}{32}
$