b^2+10b-14=-11

Solution for b^2+10b-14=-11 equation:

b^2+10b-14=-11

We move all terms to the left:

b^2+10b-14-(-11)=0

We add all the numbers together, and all the variables

b^2+10b-3=0

a = 1; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·1·(-3)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{7}}{2*1}=\frac{-10-4\sqrt{7}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{7}}{2*1}=\frac{-10+4\sqrt{7}}{2}
$