4t^2-11t+6=0

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Solution for 4t^2-11t+6=0 equation: 



4t^2-11t+6=0

a = 4; b = -11; c = +6;
Δ = b2-4ac
Δ = -112-4·4·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*4}=\frac{6}{8}
=3/4
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*4}=\frac{16}{8}
=2
$

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