0=t^2-10t+21

Solution for 0=t^2-10t+21 equation:

0=t^2-10t+21

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-t^2+10t-21=0

We add all the numbers together, and all the variables

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

a = -1; b = 10; c = -21;
Δ = b2-4ac
Δ = 102-4·(-1)·(-21)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4}{2*-1}=\frac{-14}{-2}
=+7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4}{2*-1}=\frac{-6}{-2}
=+3
$