6t^2+48t+72=0

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Solution for 6t^2+48t+72=0 equation:



6t^2+48t+72=0
a = 6; b = 48; c = +72;
Δ = b2-4ac
Δ = 482-4·6·72
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-24}{2*6}=\frac{-72}{12} =-6 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+24}{2*6}=\frac{-24}{12} =-2 $

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