Solve left(1--3/2right)x^2+x+1-k=0

$$\left( 1- \frac{ -3 }{ 2 } \right) { x }^{ 2 } +x+1-k = 0$$

Solve for k

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

Steps for Solving Linear Equation

Fraction \frac{-3}{2} can be rewritten as -\frac{3}{2} by extracting the negative sign.

$$\left(1-\left(-\frac{3}{2}\right)\right)x^{2}+x+1-k=0$$

The opposite of -\frac{3}{2} is \frac{3}{2}.

$$\left(1+\frac{3}{2}\right)x^{2}+x+1-k=0$$

Add 1 and \frac{3}{2} to get \frac{5}{2}.

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

Subtract \frac{5}{2}x^{2} from both sides. Anything subtracted from zero gives its negation.

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

Subtract x from both sides.

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

Subtract 1 from both sides.

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

The equation is in standard form.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

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

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

Solve for x (complex solution)

$$x=\frac{\sqrt{10k-9}-1}{5}$$

$$x=\frac{-\sqrt{10k-9}-1}{5}$$

Solve for x

$$x=\frac{\sqrt{10k-9}-1}{5}$$

$$x=\frac{-\sqrt{10k-9}-1}{5}\text{, }k\geq \frac{9}{10}$$