Solve 5-x_0^2=2x_{0}(3-x_{0})

$$5 - x _ { 0 } ^ { 2 } = 2 x _ { 0 } ( 3 - x _ { 0 } )$$

Solve for x_0

$$x_{0}=1$$

$$x_{0}=5$$

Steps Using Factoring

Use the distributive property to multiply 2x_{0} by 3-x_{0}.

$$5-x_{0}^{2}=6x_{0}-2x_{0}^{2}$$

Subtract 6x_{0} from both sides.

$$5-x_{0}^{2}-6x_{0}=-2x_{0}^{2}$$

Add 2x_{0}^{2} to both sides.

$$5-x_{0}^{2}-6x_{0}+2x_{0}^{2}=0$$

Combine -x_{0}^{2} and 2x_{0}^{2} to get x_{0}^{2}.

$$5+x_{0}^{2}-6x_{0}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x_{0}^{2}-6x_{0}+5=0$$

To solve the equation, factor x_{0}^{2}-6x_{0}+5 using formula x_{0}^{2}+\left(a+b\right)x_{0}+ab=\left(x_{0}+a\right)\left(x_{0}+b\right). To find a and b, set up a system to be solved.

$$a+b=-6 ab=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

Rewrite factored expression \left(x_{0}+a\right)\left(x_{0}+b\right) using the obtained values.

$$\left(x_{0}-5\right)\left(x_{0}-1\right)$$

To find equation solutions, solve x_{0}-5=0 and x_{0}-1=0.

$$x_{0}=5 x_{0}=1$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2x_{0} by 3-x_{0}.

$$5-x_{0}^{2}=6x_{0}-2x_{0}^{2}$$

Subtract 6x_{0} from both sides.

$$5-x_{0}^{2}-6x_{0}=-2x_{0}^{2}$$

Add 2x_{0}^{2} to both sides.

$$5-x_{0}^{2}-6x_{0}+2x_{0}^{2}=0$$

Combine -x_{0}^{2} and 2x_{0}^{2} to get x_{0}^{2}.

$$5+x_{0}^{2}-6x_{0}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x_{0}^{2}-6x_{0}+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x_{0}^{2}+ax_{0}+bx_{0}+5. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 5=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

Rewrite x_{0}^{2}-6x_{0}+5 as \left(x_{0}^{2}-5x_{0}\right)+\left(-x_{0}+5\right).

$$\left(x_{0}^{2}-5x_{0}\right)+\left(-x_{0}+5\right)$$

Factor out x_{0} in the first and -1 in the second group.

$$x_{0}\left(x_{0}-5\right)-\left(x_{0}-5\right)$$

Factor out common term x_{0}-5 by using distributive property.

$$\left(x_{0}-5\right)\left(x_{0}-1\right)$$

To find equation solutions, solve x_{0}-5=0 and x_{0}-1=0.

$$x_{0}=5 x_{0}=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x_{0} by 3-x_{0}.

$$5-x_{0}^{2}=6x_{0}-2x_{0}^{2}$$

Subtract 6x_{0} from both sides.

$$5-x_{0}^{2}-6x_{0}=-2x_{0}^{2}$$

Add 2x_{0}^{2} to both sides.

$$5-x_{0}^{2}-6x_{0}+2x_{0}^{2}=0$$

Combine -x_{0}^{2} and 2x_{0}^{2} to get x_{0}^{2}.

$$5+x_{0}^{2}-6x_{0}=0$$

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.

$$x_{0}^{2}-6x_{0}+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x_{0}=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5}}{2}$$

Square -6.

$$x_{0}=\frac{-\left(-6\right)±\sqrt{36-4\times 5}}{2}$$

Multiply -4 times 5.

$$x_{0}=\frac{-\left(-6\right)±\sqrt{36-20}}{2}$$

Add 36 to -20.

$$x_{0}=\frac{-\left(-6\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$x_{0}=\frac{-\left(-6\right)±4}{2}$$

The opposite of -6 is 6.

$$x_{0}=\frac{6±4}{2}$$

Now solve the equation x_{0}=\frac{6±4}{2} when ± is plus. Add 6 to 4.

$$x_{0}=\frac{10}{2}$$

Divide 10 by 2.

$$x_{0}=5$$

Now solve the equation x_{0}=\frac{6±4}{2} when ± is minus. Subtract 4 from 6.

$$x_{0}=\frac{2}{2}$$

Divide 2 by 2.

$$x_{0}=1$$

The equation is now solved.

$$x_{0}=5 x_{0}=1$$

Steps for Completing the Square

Use the distributive property to multiply 2x_{0} by 3-x_{0}.

$$5-x_{0}^{2}=6x_{0}-2x_{0}^{2}$$

Subtract 6x_{0} from both sides.

$$5-x_{0}^{2}-6x_{0}=-2x_{0}^{2}$$

Add 2x_{0}^{2} to both sides.

$$5-x_{0}^{2}-6x_{0}+2x_{0}^{2}=0$$

Combine -x_{0}^{2} and 2x_{0}^{2} to get x_{0}^{2}.

$$5+x_{0}^{2}-6x_{0}=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$x_{0}^{2}-6x_{0}=-5$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x_{0}^{2}-6x_{0}+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$x_{0}^{2}-6x_{0}+9=-5+9$$

Add -5 to 9.

$$x_{0}^{2}-6x_{0}+9=4$$

Factor x_{0}^{2}-6x_{0}+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x_{0}-3\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x_{0}-3\right)^{2}}=\sqrt{4}$$

Simplify.

$$x_{0}-3=2 x_{0}-3=-2$$

Add 3 to both sides of the equation.

$$x_{0}=5 x_{0}=1$$