21t-5=t^2-4

Solution for 21t-5=t^2-4 equation:

21t-5=t^2-4

We move all terms to the left:

21t-5-(t^2-4)=0

We get rid of parentheses

-t^2+21t+4-5=0

We add all the numbers together, and all the variables

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

a = -1; b = 21; c = -1;
Δ = b2-4ac
Δ = 212-4·(-1)·(-1)
Δ = 437
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{-(21)-\sqrt{437}}{2*-1}=\frac{-21-\sqrt{437}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{437}}{2*-1}=\frac{-21+\sqrt{437}}{-2}
$