3q^2-16q+13=0

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

3q^2-16q+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:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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