Solve 2x^2+(k-5)x-18=0

$$2 x ^ { 2 } + ( k - 5 ) x - 18 = 0$$

Solve for k

$$k=-2x+5+\frac{18}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply k-5 by x.

$$2x^{2}+kx-5x-18=0$$

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

$$kx-5x-18=-2x^{2}$$

Add 5x to both sides.

$$kx-18=-2x^{2}+5x$$

Add 18 to both sides.

$$kx=-2x^{2}+5x+18$$

The equation is in standard form.

$$xk=18+5x-2x^{2}$$

Divide both sides by x.

$$\frac{xk}{x}=-\frac{\left(2x-9\right)\left(x+2\right)}{x}$$

Dividing by x undoes the multiplication by x.

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

Divide -\left(-9+2x\right)\left(2+x\right) by x.

$$k=-2x+5+\frac{18}{x}$$

Solve for x

$$x=\frac{\sqrt{k^{2}-10k+169}-k+5}{4}$$

$$x=\frac{-\sqrt{k^{2}-10k+169}-k+5}{4}$$