b^2+6b-8=0

Solution for b^2+6b-8=0 equation:

b^2+6b-8=0

a = 1; b = 6; c = -8;
Δ = b2-4ac
Δ = 62-4·1·(-8)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{17}}{2*1}=\frac{-6-2\sqrt{17}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{17}}{2*1}=\frac{-6+2\sqrt{17}}{2}
$