10k^2+12k-94=0

Solution for 10k^2+12k-94=0 equation:

10k^2+12k-94=0

a = 10; b = 12; c = -94;
Δ = b2-4ac
Δ = 122-4·10·(-94)
Δ = 3904
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3904}=\sqrt{64*61}=\sqrt{64}*\sqrt{61}=8\sqrt{61}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{61}}{2*10}=\frac{-12-8\sqrt{61}}{20}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{61}}{2*10}=\frac{-12+8\sqrt{61}}{20}
$