10(t)=38-16(t)^2

Solution for 10(t)=38-16(t)^2 equation:

10(t)=38-16(t)^2

We move all terms to the left:

10(t)-(38-16(t)^2)=0

determiningTheFunctionDomain
-(38-16t^2)+10t=0

We get rid of parentheses

16t^2+10t-38=0

a = 16; b = 10; c = -38;
Δ = b2-4ac
Δ = 102-4·16·(-38)
Δ = 2532
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:

$\sqrt{\Delta}=\sqrt{2532}=\sqrt{4*633}=\sqrt{4}*\sqrt{633}=2\sqrt{633}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{633}}{2*16}=\frac{-10-2\sqrt{633}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{633}}{2*16}=\frac{-10+2\sqrt{633}}{32}
$