3t^2+24t+45=0

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Solution for 3t^2+24t+45=0 equation: 



3t^2+24t+45=0

a = 3; b = 24; c = +45;
Δ = b2-4ac
Δ = 242-4·3·45
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6}{2*3}=\frac{-30}{6}
=-5
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6}{2*3}=\frac{-18}{6}
=-3
$

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