b^2+6b=)

Solution for b^2+6b=) equation:

b^2+6b=)

We move all terms to the left:

b^2+6b-())=0

We add all the numbers together, and all the variables

b^2+6b=0

a = 1; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·1·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*1}=\frac{-12}{2}
=-6
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*1}=\frac{0}{2}
=0
$