Solve 5-2x(x-1)=4(3-x)-2x

$$5-2x(x-1)=4(3-x)-2x$$

Solve for x

$$x=\frac{\sqrt{2}}{2}+2\approx 2.707106781$$

$$x=-\frac{\sqrt{2}}{2}+2\approx 1.292893219$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4 by 3-x.

$$5-2x\left(x-1\right)=12-4x-2x$$

Combine -4x and -2x to get -6x.

$$5-2x\left(x-1\right)=12-6x$$

Subtract 12 from both sides.

$$5-2x\left(x-1\right)-12=-6x$$

Add 6x to both sides.

$$5-2x\left(x-1\right)-12+6x=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$5-2x\left(x-1\right)+6x=12$$

Subtract 12 from both sides.

$$5-2x\left(x-1\right)+6x-12=0$$

Use the distributive property to multiply -2x by x-1.

$$5-2x^{2}+2x+6x-12=0$$

Combine 2x and 6x to get 8x.

$$5-2x^{2}+8x-12=0$$

Subtract 12 from 5 to get -7.

$$-7-2x^{2}+8x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2x^{2}+8x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 8 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-2\right)\left(-7\right)}}{2\left(-2\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-2\right)\left(-7\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-8±\sqrt{64+8\left(-7\right)}}{2\left(-2\right)}$$

Multiply 8 times -7.

$$x=\frac{-8±\sqrt{64-56}}{2\left(-2\right)}$$

Add 64 to -56.

$$x=\frac{-8±\sqrt{8}}{2\left(-2\right)}$$

Take the square root of 8.

$$x=\frac{-8±2\sqrt{2}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-8±2\sqrt{2}}{-4}$$

Now solve the equation x=\frac{-8±2\sqrt{2}}{-4} when ± is plus. Add -8 to 2\sqrt{2}.

$$x=\frac{2\sqrt{2}-8}{-4}$$

Divide 2\sqrt{2}-8 by -4.

$$x=-\frac{\sqrt{2}}{2}+2$$

Now solve the equation x=\frac{-8±2\sqrt{2}}{-4} when ± is minus. Subtract 2\sqrt{2} from -8.

$$x=\frac{-2\sqrt{2}-8}{-4}$$

Divide -8-2\sqrt{2} by -4.

$$x=\frac{\sqrt{2}}{2}+2$$

The equation is now solved.

$$x=-\frac{\sqrt{2}}{2}+2 x=\frac{\sqrt{2}}{2}+2$$

Steps for Completing the Square

Use the distributive property to multiply 4 by 3-x.

$$5-2x\left(x-1\right)=12-4x-2x$$

Combine -4x and -2x to get -6x.

$$5-2x\left(x-1\right)=12-6x$$

Add 6x to both sides.

$$5-2x\left(x-1\right)+6x=12$$

Use the distributive property to multiply -2x by x-1.

$$5-2x^{2}+2x+6x=12$$

Combine 2x and 6x to get 8x.

$$5-2x^{2}+8x=12$$

Subtract 5 from both sides.

$$-2x^{2}+8x=12-5$$

Subtract 5 from 12 to get 7.

$$-2x^{2}+8x=7$$

Divide both sides by -2.

$$\frac{-2x^{2}+8x}{-2}=\frac{7}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{8}{-2}x=\frac{7}{-2}$$

Divide 8 by -2.

$$x^{2}-4x=\frac{7}{-2}$$

Divide 7 by -2.

$$x^{2}-4x=-\frac{7}{2}$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=-\frac{7}{2}+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-\frac{7}{2}+4$$

Add -\frac{7}{2} to 4.

$$x^{2}-4x+4=\frac{1}{2}$$

Factor x^{2}-4x+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}.

$$\left(x-2\right)^{2}=\frac{1}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{1}{2}}$$

Simplify.

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

Add 2 to both sides of the equation.

$$x=\frac{\sqrt{2}}{2}+2 x=-\frac{\sqrt{2}}{2}+2$$