4t2+11t+2=0

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Solution for 4t2+11t+2=0 equation: 



4t^2+11t+2=0

a = 4; b = 11; c = +2;
Δ = b2-4ac
Δ = 112-4·4·2
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{89}}{2*4}=\frac{-11-\sqrt{89}}{8}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{89}}{2*4}=\frac{-11+\sqrt{89}}{8}
$

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