Solve (4+m)^2-4x(5m-5)=0

$$( 4 + m ) ^ { 2 } - 4 x ( 5 m - 5 ) = 0$$

Solve for x

$$x=-\frac{\left(m+4\right)^{2}}{20\left(1-m\right)}$$

$$m\neq 1$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+m\right)^{2}.

$$16+8m+m^{2}-4x\left(5m-5\right)=0$$

Use the distributive property to multiply -4x by 5m-5.

$$16+8m+m^{2}-20mx+20x=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$8m+m^{2}-20mx+20x=-16$$

Subtract 8m from both sides.

$$m^{2}-20mx+20x=-16-8m$$

Subtract m^{2} from both sides.

$$-20mx+20x=-16-8m-m^{2}$$

Combine all terms containing x.

$$\left(-20m+20\right)x=-16-8m-m^{2}$$

The equation is in standard form.

$$\left(20-20m\right)x=-m^{2}-8m-16$$

Divide both sides by 20-20m.

$$\frac{\left(20-20m\right)x}{20-20m}=-\frac{\left(m+4\right)^{2}}{20-20m}$$

Dividing by 20-20m undoes the multiplication by 20-20m.

$$x=-\frac{\left(m+4\right)^{2}}{20-20m}$$

Divide -\left(m+4\right)^{2} by 20-20m.

$$x=-\frac{\left(m+4\right)^{2}}{20\left(1-m\right)}$$

Solve for m (complex solution)

$$m=2\left(5\sqrt{x\left(x-1\right)}+5x-2\right)$$

$$m=2\left(-5\sqrt{x\left(x-1\right)}+5x-2\right)$$

Solve for m

$$m=2\left(5\sqrt{x\left(x-1\right)}+5x-2\right)$$

$$m=2\left(-5\sqrt{x\left(x-1\right)}+5x-2\right)\text{, }x\geq 1\text{ or }x\leq 0$$