t2+11t+24=0

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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 $

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