3k^2+16k-64=0

Solution for 3k^2+16k-64=0 equation:

3k^2+16k-64=0

a = 3; b = 16; c = -64;
Δ = b2-4ac
Δ = 162-4·3·(-64)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-32}{2*3}=\frac{-48}{6}
=-8
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+32}{2*3}=\frac{16}{6}
=2+2/3
$