0=5b(b-7)-4(3+2b)

Solution for 0=5b(b-7)-4(3+2b) equation:

0=5b(b-7)-4(3+2b)

We move all terms to the left:

0-(5b(b-7)-4(3+2b))=0

We add all the numbers together, and all the variables

-(5b(b-7)-4(2b+3))+0=0

We add all the numbers together, and all the variables

-(5b(b-7)-4(2b+3))=0


We calculate terms in parentheses: -(5b(b-7)-4(2b+3)), so:

5b(b-7)-4(2b+3)

We multiply parentheses

5b^2-35b-8b-12

We add all the numbers together, and all the variables

5b^2-43b-12

Back to the equation:

-(5b^2-43b-12)


We get rid of parentheses

-5b^2+43b+12=0

a = -5; b = 43; c = +12;
Δ = b2-4ac
Δ = 432-4·(-5)·12
Δ = 2089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-\sqrt{2089}}{2*-5}=\frac{-43-\sqrt{2089}}{-10}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+\sqrt{2089}}{2*-5}=\frac{-43+\sqrt{2089}}{-10}
$