4t+t^2=21

Solution for 4t+t^2=21 equation:

4t+t^2=21

We move all terms to the left:

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

a = 1; b = 4; c = -21;
Δ = b2-4ac
Δ = 42-4·1·(-21)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-10}{2*1}=\frac{-14}{2}
=-7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+10}{2*1}=\frac{6}{2}
=3
$