Solve (b-4)x^2+(2b-8)x+15=0?

$$( b - 4 ) x ^ { 2 } + ( 2 b - 8 ) x + 15 = 0 ?$$

Solve for b

$$b=-\frac{15-8x-4x^{2}}{x\left(x+2\right)}$$

$$x\neq -2\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply b-4 by x^{2}.

$$bx^{2}-4x^{2}+\left(2b-8\right)x+15=0$$

Use the distributive property to multiply 2b-8 by x.

$$bx^{2}-4x^{2}+2bx-8x+15=0$$

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

$$bx^{2}+2bx-8x+15=4x^{2}$$

Add 8x to both sides.

$$bx^{2}+2bx+15=4x^{2}+8x$$

Subtract 15 from both sides.

$$bx^{2}+2bx=4x^{2}+8x-15$$

Combine all terms containing b.

$$\left(x^{2}+2x\right)b=4x^{2}+8x-15$$

Divide both sides by x^{2}+2x.

$$\frac{\left(x^{2}+2x\right)b}{x^{2}+2x}=\frac{4x^{2}+8x-15}{x^{2}+2x}$$

Dividing by x^{2}+2x undoes the multiplication by x^{2}+2x.

$$b=\frac{4x^{2}+8x-15}{x^{2}+2x}$$

Divide 4x^{2}+8x-15 by x^{2}+2x.

$$b=\frac{4x^{2}+8x-15}{x\left(x+2\right)}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(b-19\right)\left(b-4\right)}-b+4}{b-4}$$

$$x=\frac{-\sqrt{\left(b-19\right)\left(b-4\right)}-b+4}{b-4}\text{, }b\neq 4$$

Solve for x

$$x=\frac{\sqrt{\left(b-19\right)\left(b-4\right)}-b+4}{b-4}$$

$$x=\frac{-\sqrt{\left(b-19\right)\left(b-4\right)}-b+4}{b-4}\text{, }b\geq 19\text{ or }b<4$$