5t^2+21t+22=0

Solution for 5t^2+21t+22=0 equation:

5t^2+21t+22=0

a = 5; b = 21; c = +22;
Δ = b2-4ac
Δ = 212-4·5·22
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-1}{2*5}=\frac{-22}{10}
=-2+1/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+1}{2*5}=\frac{-20}{10}
=-2
$