3t^2+47t-16=0

Solution for 3t^2+47t-16=0 equation:

3t^2+47t-16=0

a = 3; b = 47; c = -16;
Δ = b2-4ac
Δ = 472-4·3·(-16)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-49}{2*3}=\frac{-96}{6}
=-16
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+49}{2*3}=\frac{2}{6}
=1/3
$