12(b2)=8(b+5)

Solution for 12(b2)=8(b+5) equation:

12(b2)=8(b+5)

We move all terms to the left:

12(b2)-(8(b+5))=0

We add all the numbers together, and all the variables

12b^2-(8(b+5))=0


We calculate terms in parentheses: -(8(b+5)), so:

8(b+5)

We multiply parentheses

8b+40

Back to the equation:

-(8b+40)


We get rid of parentheses

12b^2-8b-40=0

a = 12; b = -8; c = -40;
Δ = b2-4ac
Δ = -82-4·12·(-40)
Δ = 1984
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{31}}{2*12}=\frac{8-8\sqrt{31}}{24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{31}}{2*12}=\frac{8+8\sqrt{31}}{24}
$