t2+11t+24=0

Solution for t2+11t+24=0 equation:

t2+11t+24=0

We add all the numbers together, and all the variables

t^2+11t+24=0

a = 1; b = 11; c = +24;
Δ = b2-4ac
Δ = 112-4·1·24
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*1}=\frac{-16}{2}
=-8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*1}=\frac{-6}{2}
=-3
$