b^2+12b-58=0

Solution for b^2+12b-58=0 equation:

b^2+12b-58=0

a = 1; b = 12; c = -58;
Δ = b2-4ac
Δ = 122-4·1·(-58)
Δ = 376
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:

$\sqrt{\Delta}=\sqrt{376}=\sqrt{4*94}=\sqrt{4}*\sqrt{94}=2\sqrt{94}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{94}}{2*1}=\frac{-12-2\sqrt{94}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{94}}{2*1}=\frac{-12+2\sqrt{94}}{2}
$