0(t)=-16t2+90t+3

Solution for 0(t)=-16t2+90t+3 equation:

0(t)=-16t^2+90t+3

We move all terms to the left:

0(t)-(-16t^2+90t+3)=0

We add all the numbers together, and all the variables

-(-16t^2+90t+3)+t=0

We get rid of parentheses

16t^2-90t+t-3=0

We add all the numbers together, and all the variables

16t^2-89t-3=0

a = 16; b = -89; c = -3;
Δ = b2-4ac
Δ = -892-4·16·(-3)
Δ = 8113
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-89)-\sqrt{8113}}{2*16}=\frac{89-\sqrt{8113}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-89)+\sqrt{8113}}{2*16}=\frac{89+\sqrt{8113}}{32}
$