Solve 16=1+20x-5x^2

$$16 = 1 + 20 x - 5 x ^ { 2 }$$

Solve for x

$$x=1$$

$$x=3$$

Steps Using Factoring By Grouping

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

$$1+20x-5x^{2}=16$$

Subtract 16 from both sides.

$$1+20x-5x^{2}-16=0$$

Subtract 16 from 1 to get -15.

$$-15+20x-5x^{2}=0$$

Divide both sides by 5.

$$-3+4x-x^{2}=0$$

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

$$-x^{2}+4x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=4 ab=-\left(-3\right)=3$$

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.

$$a=3 b=1$$

Rewrite -x^{2}+4x-3 as \left(-x^{2}+3x\right)+\left(x-3\right).

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

Factor out -x in -x^{2}+3x.

$$-x\left(x-3\right)+x-3$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x+1\right)$$

To find equation solutions, solve x-3=0 and -x+1=0.

$$x=3 x=1$$

Steps Using the Quadratic Formula

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

$$1+20x-5x^{2}=16$$

Subtract 16 from both sides.

$$1+20x-5x^{2}-16=0$$

Subtract 16 from 1 to get -15.

$$-15+20x-5x^{2}=0$$

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.

$$-5x^{2}+20x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 20 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-5\right)\left(-15\right)}}{2\left(-5\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-5\right)\left(-15\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-20±\sqrt{400+20\left(-15\right)}}{2\left(-5\right)}$$

Multiply 20 times -15.

$$x=\frac{-20±\sqrt{400-300}}{2\left(-5\right)}$$

Add 400 to -300.

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

Take the square root of 100.

$$x=\frac{-20±10}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-20±10}{-10}$$

Now solve the equation x=\frac{-20±10}{-10} when ± is plus. Add -20 to 10.

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

Divide -10 by -10.

$$x=1$$

Now solve the equation x=\frac{-20±10}{-10} when ± is minus. Subtract 10 from -20.

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

Divide -30 by -10.

$$x=3$$

The equation is now solved.

$$x=1 x=3$$

Steps for Completing the Square

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

$$1+20x-5x^{2}=16$$

Subtract 1 from both sides.

$$20x-5x^{2}=16-1$$

Subtract 1 from 16 to get 15.

$$20x-5x^{2}=15$$

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.

$$-5x^{2}+20x=15$$

Divide both sides by -5.

$$\frac{-5x^{2}+20x}{-5}=\frac{15}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{20}{-5}x=\frac{15}{-5}$$

Divide 20 by -5.

$$x^{2}-4x=\frac{15}{-5}$$

Divide 15 by -5.

$$x^{2}-4x=-3$$

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.

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

Square -2.

$$x^{2}-4x+4=-3+4$$

Add -3 to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$