-16t^2+11t+3=0

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Solution for -16t^2+11t+3=0 equation: 



-16t^2+11t+3=0

a = -16; b = 11; c = +3;
Δ = b2-4ac
Δ = 112-4·(-16)·3
Δ = 313
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{313}}{2*-16}=\frac{-11-\sqrt{313}}{-32}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{313}}{2*-16}=\frac{-11+\sqrt{313}}{-32}
$

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