3q^2+-10q+-77=0

Solution for 3q^2+-10q+-77=0 equation:

3q^2+-10q+-77=0

We add all the numbers together, and all the variables

3q^2-10q=0

a = 3; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·3·0
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{-(-10)-10}{2*3}=\frac{0}{6}
=0
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*3}=\frac{20}{6}
=3+1/3
$