b^2+6b-2=10

Solution for b^2+6b-2=10 equation:

b^2+6b-2=10

We move all terms to the left:

b^2+6b-2-(10)=0

We add all the numbers together, and all the variables

b^2+6b-12=0

a = 1; b = 6; c = -12;
Δ = b2-4ac
Δ = 62-4·1·(-12)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{21}}{2*1}=\frac{-6-2\sqrt{21}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{21}}{2*1}=\frac{-6+2\sqrt{21}}{2}
$