-7k^2+12k+8=0

Solution for -7k^2+12k+8=0 equation:

-7k^2+12k+8=0

a = -7; b = 12; c = +8;
Δ = b2-4ac
Δ = 122-4·(-7)·8
Δ = 368
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:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{23}}{2*-7}=\frac{-12-4\sqrt{23}}{-14}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{23}}{2*-7}=\frac{-12+4\sqrt{23}}{-14}
$