Solve 10=5+10x-5x^2

$$10 = 5 + 10 x - 5 x ^ { 2 }$$

Solve for x

$$x=1$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$5+10x-5x^{2}=10$$

Subtract 10 from both sides.

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

Subtract 10 from 5 to get -5.

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

Divide both sides by 5.

$$-1+2x-x^{2}=0$$

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

$$-x^{2}+2x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=2 ab=-\left(-1\right)=1$$

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

$$a=1 b=1$$

Rewrite -x^{2}+2x-1 as \left(-x^{2}+x\right)+\left(x-1\right).

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

Factor out -x in -x^{2}+x.

$$-x\left(x-1\right)+x-1$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(-x+1\right)$$

To find equation solutions, solve x-1=0 and -x+1=0.

$$x=1 x=1$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$5+10x-5x^{2}=10$$

Subtract 10 from both sides.

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

Subtract 10 from 5 to get -5.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 10 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-5\right)\left(-5\right)}}{2\left(-5\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-5\right)\left(-5\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-10±\sqrt{100+20\left(-5\right)}}{2\left(-5\right)}$$

Multiply 20 times -5.

$$x=\frac{-10±\sqrt{100-100}}{2\left(-5\right)}$$

Add 100 to -100.

$$x=\frac{-10±\sqrt{0}}{2\left(-5\right)}$$

Take the square root of 0.

$$x=-\frac{10}{2\left(-5\right)}$$

Multiply 2 times -5.

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

Divide -10 by -10.

$$x=1$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$5+10x-5x^{2}=10$$

Subtract 5 from both sides.

$$10x-5x^{2}=10-5$$

Subtract 5 from 10 to get 5.

$$10x-5x^{2}=5$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-5x^{2}+10x=5$$

Divide both sides by -5.

$$\frac{-5x^{2}+10x}{-5}=\frac{5}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{10}{-5}x=\frac{5}{-5}$$

Divide 10 by -5.

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

Divide 5 by -5.

$$x^{2}-2x=-1$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=-1+1$$

Add -1 to 1.

$$x^{2}-2x+1=0$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-1=0 x-1=0$$

Add 1 to both sides of the equation.

$$x=1 x=1$$

The equation is now solved. Solutions are the same.

$$x=1$$