b^2+6b-16=0b=

Solution for b^2+6b-16=0b= equation:

b^2+6b-16=0b=

We move all terms to the left:

b^2+6b-16-(0b)=0

We add all the numbers together, and all the variables

b^2+5b-16=0

a = 1; b = 5; c = -16;
Δ = b2-4ac
Δ = 52-4·1·(-16)
Δ = 89
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{-(5)-\sqrt{89}}{2*1}=\frac{-5-\sqrt{89}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{89}}{2*1}=\frac{-5+\sqrt{89}}{2}
$