14b+6=2b(b-3)

Solution for 14b+6=2b(b-3) equation:

14b+6=2b(b-3)

We move all terms to the left:

14b+6-(2b(b-3))=0


We calculate terms in parentheses: -(2b(b-3)), so:

2b(b-3)

We multiply parentheses

2b^2-6b

Back to the equation:

-(2b^2-6b)


We get rid of parentheses

-2b^2+14b+6b+6=0

We add all the numbers together, and all the variables

-2b^2+20b+6=0

a = -2; b = 20; c = +6;
Δ = b2-4ac
Δ = 202-4·(-2)·6
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{7}}{2*-2}=\frac{-20-8\sqrt{7}}{-4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{7}}{2*-2}=\frac{-20+8\sqrt{7}}{-4}
$