6k^2+16k-7=0

Solution for 6k^2+16k-7=0 equation:

6k^2+16k-7=0

a = 6; b = 16; c = -7;
Δ = b2-4ac
Δ = 162-4·6·(-7)
Δ = 424
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:

$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{106}}{2*6}=\frac{-16-2\sqrt{106}}{12}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{106}}{2*6}=\frac{-16+2\sqrt{106}}{12}
$