Solve left(1-9/4right)x^2+x+1-k=0

$$\left( 1- \frac{ 9 }{ 4 } \right) { x }^{ 2 } +x+1-k = 0$$

Solve for k

$$k=-\frac{5x^{2}}{4}+x+1$$

Steps for Solving Linear Equation

Subtract \frac{9}{4} from 1 to get -\frac{5}{4}.

$$-\frac{5}{4}x^{2}+x+1-k=0$$

Add \frac{5}{4}x^{2} to both sides. Anything plus zero gives itself.

$$x+1-k=\frac{5}{4}x^{2}$$

Subtract x from both sides.

$$1-k=\frac{5}{4}x^{2}-x$$

Subtract 1 from both sides.

$$-k=\frac{5}{4}x^{2}-x-1$$

The equation is in standard form.

$$-k=\frac{5x^{2}}{4}-x-1$$

Divide both sides by -1.

$$\frac{-k}{-1}=\frac{\frac{5x^{2}}{4}-x-1}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$k=\frac{\frac{5x^{2}}{4}-x-1}{-1}$$

Divide \frac{5x^{2}}{4}-x-1 by -1.

$$k=-\frac{5x^{2}}{4}+x+1$$

Solve for x (complex solution)

$$x=\frac{-2\sqrt{6-5k}+2}{5}$$

$$x=\frac{2\sqrt{6-5k}+2}{5}$$

Solve for x

$$x=\frac{-2\sqrt{6-5k}+2}{5}$$

$$x=\frac{2\sqrt{6-5k}+2}{5}\text{, }k\leq \frac{6}{5}$$