403=-16t^2+75t+403

Solution for 403=-16t^2+75t+403 equation:

403=-16t^2+75t+403

We move all terms to the left:

403-(-16t^2+75t+403)=0

We get rid of parentheses

16t^2-75t-403+403=0

We add all the numbers together, and all the variables

16t^2-75t=0

a = 16; b = -75; c = 0;
Δ = b2-4ac
Δ = -752-4·16·0
Δ = 5625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5625}=75$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-75)-75}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-75)+75}{2*16}=\frac{150}{32}
=4+11/16
$