(2b+1)(3b-4)=

Solution for (2b+1)(3b-4)= equation:

(2b+1)(3b-4)=

We move all terms to the left:

(2b+1)(3b-4)-()=0

We add all the numbers together, and all the variables

(2b+1)(3b-4)=0

We multiply parentheses ..

(+6b^2-8b+3b-4)=0

We get rid of parentheses

6b^2-8b+3b-4=0

We add all the numbers together, and all the variables

6b^2-5b-4=0

a = 6; b = -5; c = -4;
Δ = b2-4ac
Δ = -52-4·6·(-4)
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{-(-5)-11}{2*6}=\frac{-6}{12}
=-1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*6}=\frac{16}{12}
=1+1/3
$