t2+11t+10=0

Solution for t2+11t+10=0 equation:

t2+11t+10=0

We add all the numbers together, and all the variables

t^2+11t+10=0

a = 1; b = 11; c = +10;
Δ = b2-4ac
Δ = 112-4·1·10
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-9}{2*1}=\frac{-20}{2}
=-10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+9}{2*1}=\frac{-2}{2}
=-1
$