3q^2-14q+16=0

Solution for 3q^2-14q+16=0 equation:

3q^2-14q+16=0

a = 3; b = -14; c = +16;
Δ = b2-4ac
Δ = -142-4·3·16
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2}{2*3}=\frac{12}{6}
=2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2}{2*3}=\frac{16}{6}
=2+2/3
$