(4b+5)(b-2)=0

Solution for (4b+5)(b-2)=0 equation:

(4b+5)(b-2)=0

We multiply parentheses ..

(+4b^2-8b+5b-10)=0

We get rid of parentheses

4b^2-8b+5b-10=0

We add all the numbers together, and all the variables

4b^2-3b-10=0

a = 4; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·4·(-10)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*4}=\frac{-10}{8}
=-1+1/4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*4}=\frac{16}{8}
=2
$