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

Solution for (2b-3)(b+4)=0 equation:

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

We multiply parentheses ..

(+2b^2+8b-3b-12)=0

We get rid of parentheses

2b^2+8b-3b-12=0

We add all the numbers together, and all the variables

2b^2+5b-12=0

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