5t^2+12t+4=0

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Solution for 5t^2+12t+4=0 equation: 



5t^2+12t+4=0

a = 5; b = 12; c = +4;
Δ = b2-4ac
Δ = 122-4·5·4
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*5}=\frac{-20}{10}
=-2
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*5}=\frac{-4}{10}
=-2/5
$

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