3k^2+11k-4=11

Solution for 3k^2+11k-4=11 equation:

3k^2+11k-4=11

We move all terms to the left:

3k^2+11k-4-(11)=0

We add all the numbers together, and all the variables

3k^2+11k-15=0

a = 3; b = 11; c = -15;
Δ = b2-4ac
Δ = 112-4·3·(-15)
Δ = 301
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{301}}{2*3}=\frac{-11-\sqrt{301}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{301}}{2*3}=\frac{-11+\sqrt{301}}{6}
$