3q^2+10q-1400=0

Solution for 3q^2+10q-1400=0 equation:

3q^2+10q-1400=0

a = 3; b = 10; c = -1400;
Δ = b2-4ac
Δ = 102-4·3·(-1400)
Δ = 16900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16900}=130$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-130}{2*3}=\frac{-140}{6}
=-23+1/3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+130}{2*3}=\frac{120}{6}
=20
$