3k^2+7k-40=0

Solution for 3k^2+7k-40=0 equation:

3k^2+7k-40=0

a = 3; b = 7; c = -40;
Δ = b2-4ac
Δ = 72-4·3·(-40)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-23}{2*3}=\frac{-30}{6}
=-5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*3}=\frac{16}{6}
=2+2/3
$