r^2+12r-3=0

Solution for r^2+12r-3=0 equation:

r^2+12r-3=0

a = 1; b = 12; c = -3;
Δ = b2-4ac
Δ = 122-4·1·(-3)
Δ = 156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{39}}{2*1}=\frac{-12-2\sqrt{39}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{39}}{2*1}=\frac{-12+2\sqrt{39}}{2}
$