0(t)=-16t^2+150t+30

Solution for 0(t)=-16t^2+150t+30 equation:

0(t)=-16t^2+150t+30

We move all terms to the left:

0(t)-(-16t^2+150t+30)=0

We add all the numbers together, and all the variables

-(-16t^2+150t+30)+t=0

We get rid of parentheses

16t^2-150t+t-30=0

We add all the numbers together, and all the variables

16t^2-149t-30=0

a = 16; b = -149; c = -30;
Δ = b2-4ac
Δ = -1492-4·16·(-30)
Δ = 24121
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{-(-149)-\sqrt{24121}}{2*16}=\frac{149-\sqrt{24121}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-149)+\sqrt{24121}}{2*16}=\frac{149+\sqrt{24121}}{32}
$