21=-t2+10t+5

Solution for 21=-t2+10t+5 equation:

21=-t2+10t+5

We move all terms to the left:

21-(-t2+10t+5)=0

We add all the numbers together, and all the variables

-(-1t^2+10t+5)+21=0

We get rid of parentheses

1t^2-10t-5+21=0

We add all the numbers together, and all the variables

t^2-10t+16=0

a = 1; b = -10; c = +16;
Δ = b2-4ac
Δ = -102-4·1·16
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6}{2*1}=\frac{4}{2}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6}{2*1}=\frac{16}{2}
=8
$