35=-16x^2+53x+3

Solution for 35=-16x^2+53x+3 equation:

35=-16x^2+53x+3

We move all terms to the left:

35-(-16x^2+53x+3)=0

We get rid of parentheses

16x^2-53x-3+35=0

We add all the numbers together, and all the variables

16x^2-53x+32=0

a = 16; b = -53; c = +32;
Δ = b2-4ac
Δ = -532-4·16·32
Δ = 761
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{-(-53)-\sqrt{761}}{2*16}=\frac{53-\sqrt{761}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+\sqrt{761}}{2*16}=\frac{53+\sqrt{761}}{32}
$