3b^2+7b+-6=0

Solution for 3b^2+7b+-6=0 equation:

3b^2+7b+-6=0

We add all the numbers together, and all the variables

3b^2+7b=0

a = 3; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·3·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*3}=\frac{-14}{6}
=-2+1/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*3}=\frac{0}{6}
=0
$