a^2+12a-67=0

Solution for a^2+12a-67=0 equation:

a^2+12a-67=0

a = 1; b = 12; c = -67;
Δ = b2-4ac
Δ = 122-4·1·(-67)
Δ = 412
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{103}}{2*1}=\frac{-12-2\sqrt{103}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{103}}{2*1}=\frac{-12+2\sqrt{103}}{2}
$