Solve 4x^2-4x+1^2+x^2-10x-5-x+4=0

$$4 { x }^{ 2 } -4x+ { 1 }^{ 2 } + { x }^{ 2 } -10x-5-x+4=0$$

Solve for x

$$x=3$$

$$x=0$$

Steps Using Factoring

Calculate 1 to the power of 2 and get 1.

$$4x^{2}-4x+1+x^{2}-10x-5-x+4=0$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-4x+1-10x-5-x+4=0$$

Combine -4x and -10x to get -14x.

$$5x^{2}-14x+1-5-x+4=0$$

Subtract 5 from 1 to get -4.

$$5x^{2}-14x-4-x+4=0$$

Combine -14x and -x to get -15x.

$$5x^{2}-15x-4+4=0$$

Add -4 and 4 to get 0.

$$5x^{2}-15x=0$$

Factor out x.

$$x\left(5x-15\right)=0$$

To find equation solutions, solve x=0 and 5x-15=0.

$$x=0 x=3$$

Steps Using the Quadratic Formula

Calculate 1 to the power of 2 and get 1.

$$4x^{2}-4x+1+x^{2}-10x-5-x+4=0$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-4x+1-10x-5-x+4=0$$

Combine -4x and -10x to get -14x.

$$5x^{2}-14x+1-5-x+4=0$$

Subtract 5 from 1 to get -4.

$$5x^{2}-14x-4-x+4=0$$

Combine -14x and -x to get -15x.

$$5x^{2}-15x-4+4=0$$

Add -4 and 4 to get 0.

$$5x^{2}-15x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -15 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}}}{2\times 5}$$

Take the square root of \left(-15\right)^{2}.

$$x=\frac{-\left(-15\right)±15}{2\times 5}$$

The opposite of -15 is 15.

$$x=\frac{15±15}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{15±15}{10}$$

Now solve the equation x=\frac{15±15}{10} when ± is plus. Add 15 to 15.

$$x=\frac{30}{10}$$

Divide 30 by 10.

$$x=3$$

Now solve the equation x=\frac{15±15}{10} when ± is minus. Subtract 15 from 15.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

The equation is now solved.

$$x=3 x=0$$

Steps for Completing the Square

Calculate 1 to the power of 2 and get 1.

$$4x^{2}-4x+1+x^{2}-10x-5-x+4=0$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-4x+1-10x-5-x+4=0$$

Combine -4x and -10x to get -14x.

$$5x^{2}-14x+1-5-x+4=0$$

Subtract 5 from 1 to get -4.

$$5x^{2}-14x-4-x+4=0$$

Combine -14x and -x to get -15x.

$$5x^{2}-15x-4+4=0$$

Add -4 and 4 to get 0.

$$5x^{2}-15x=0$$

Divide both sides by 5.

$$\frac{5x^{2}-15x}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{15}{5}\right)x=\frac{0}{5}$$

Divide -15 by 5.

$$x^{2}-3x=\frac{0}{5}$$

Divide 0 by 5.

$$x^{2}-3x=0$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{9}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{3}{2} x-\frac{3}{2}=-\frac{3}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=3 x=0$$