3q^2+10q-16=0

Solution for 3q^2+10q-16=0 equation:

3q^2+10q-16=0

a = 3; b = 10; c = -16;
Δ = b2-4ac
Δ = 102-4·3·(-16)
Δ = 292
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{73}}{2*3}=\frac{-10-2\sqrt{73}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{73}}{2*3}=\frac{-10+2\sqrt{73}}{6}
$