3k^2-3k-4=7

Solution for 3k^2-3k-4=7 equation:

3k^2-3k-4=7

We move all terms to the left:

3k^2-3k-4-(7)=0

We add all the numbers together, and all the variables

3k^2-3k-11=0

a = 3; b = -3; c = -11;
Δ = b2-4ac
Δ = -32-4·3·(-11)
Δ = 141
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{-(-3)-\sqrt{141}}{2*3}=\frac{3-\sqrt{141}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{141}}{2*3}=\frac{3+\sqrt{141}}{6}
$