(t^2)+10t+24=0

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Solution for (t^2)+10t+24=0 equation: 



(t^2)+10t+24=0

a = 1; b = 10; c = +24;
Δ = b2-4ac
Δ = 102-4·1·24
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*1}=\frac{-12}{2}
=-6
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*1}=\frac{-8}{2}
=-4
$

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