3b^2+31b+70=0

Solution for 3b^2+31b+70=0 equation:

3b^2+31b+70=0

a = 3; b = 31; c = +70;
Δ = b2-4ac
Δ = 312-4·3·70
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-11}{2*3}=\frac{-42}{6}
=-7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+11}{2*3}=\frac{-20}{6}
=-3+1/3
$