Solve k^2+k-12=y

$$k ^ { 2 } + k - 12 = y$$

Solve for y

$$y=\left(k-3\right)\left(k+4\right)$$

Solution Steps

Swap sides so that all variable terms are on the left hand side.

$$y=k^{2}+k-12$$

Solve for k (complex solution)

$$k=\frac{\sqrt{4y+49}-1}{2}$$

$$k=\frac{-\sqrt{4y+49}-1}{2}$$

Solve for k

$$k=\frac{\sqrt{4y+49}-1}{2}$$

$$k=\frac{-\sqrt{4y+49}-1}{2}\text{, }y\geq -\frac{49}{4}$$