3q^2+13q+10=0

Solution for 3q^2+13q+10=0 equation:

3q^2+13q+10=0

a = 3; b = 13; c = +10;
Δ = b2-4ac
Δ = 132-4·3·10
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-7}{2*3}=\frac{-20}{6}
=-3+1/3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+7}{2*3}=\frac{-6}{6}
=-1
$