8t^2-12t+4=0

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Solution for 8t^2-12t+4=0 equation: 



8t^2-12t+4=0

a = 8; b = -12; c = +4;
Δ = b2-4ac
Δ = -122-4·8·4
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4}{2*8}=\frac{8}{16}
=1/2
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*8}=\frac{16}{16}
=1
$

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