3t^2+16t+13=0

Solution for 3t^2+16t+13=0 equation:

3t^2+16t+13=0

a = 3; b = 16; c = +13;
Δ = b2-4ac
Δ = 162-4·3·13
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-10}{2*3}=\frac{-26}{6}
=-4+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+10}{2*3}=\frac{-6}{6}
=-1
$