(6b-7)(b+4)=0

Solution for (6b-7)(b+4)=0 equation:

(6b-7)(b+4)=0

We multiply parentheses ..

(+6b^2+24b-7b-28)=0

We get rid of parentheses

6b^2+24b-7b-28=0

We add all the numbers together, and all the variables

6b^2+17b-28=0

a = 6; b = 17; c = -28;
Δ = b2-4ac
Δ = 172-4·6·(-28)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-31}{2*6}=\frac{-48}{12}
=-4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+31}{2*6}=\frac{14}{12}
=1+1/6
$