8t2+6t+1=0

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Solution for 8t2+6t+1=0 equation: 



8t^2+6t+1=0

a = 8; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·8·1
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{-(6)-2}{2*8}=\frac{-8}{16}
=-1/2
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*8}=\frac{-4}{16}
=-1/4
$

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