10(t)=-16t^2+16

Solution for 10(t)=-16t^2+16 equation:

10(t)=-16t^2+16

We move all terms to the left:

10(t)-(-16t^2+16)=0

We get rid of parentheses

16t^2+10t-16=0

a = 16; b = 10; c = -16;
Δ = b2-4ac
Δ = 102-4·16·(-16)
Δ = 1124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1124}=\sqrt{4*281}=\sqrt{4}*\sqrt{281}=2\sqrt{281}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{281}}{2*16}=\frac{-10-2\sqrt{281}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{281}}{2*16}=\frac{-10+2\sqrt{281}}{32}
$