Solve 16x+20/2=2x^2

$$\frac { 16 x + 20 } { 2 } = 2 x ^ { 2 }$$

Solve for x

$$x=-1$$

$$x=5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

$$16x+20=4x^{2}$$

Subtract 4x^{2} from both sides.

$$16x+20-4x^{2}=0$$

Divide both sides by 4.

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

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

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

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

$$a+b=4 ab=-5=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=5 b=-1$$

Rewrite -x^{2}+4x+5 as \left(-x^{2}+5x\right)+\left(-x+5\right).

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

Factor out -x in the first and -1 in the second group.

$$-x\left(x-5\right)-\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(-x-1\right)$$

To find equation solutions, solve x-5=0 and -x-1=0.

$$x=5 x=-1$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$16x+20=4x^{2}$$

Subtract 4x^{2} from both sides.

$$16x+20-4x^{2}=0$$

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.

$$-4x^{2}+16x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 16 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-4\right)\times 20}}{2\left(-4\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-4\right)\times 20}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-16±\sqrt{256+16\times 20}}{2\left(-4\right)}$$

Multiply 16 times 20.

$$x=\frac{-16±\sqrt{256+320}}{2\left(-4\right)}$$

Add 256 to 320.

$$x=\frac{-16±\sqrt{576}}{2\left(-4\right)}$$

Take the square root of 576.

$$x=\frac{-16±24}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-16±24}{-8}$$

Now solve the equation x=\frac{-16±24}{-8} when ± is plus. Add -16 to 24.

$$x=\frac{8}{-8}$$

Divide 8 by -8.

$$x=-1$$

Now solve the equation x=\frac{-16±24}{-8} when ± is minus. Subtract 24 from -16.

$$x=-\frac{40}{-8}$$

Divide -40 by -8.

$$x=5$$

The equation is now solved.

$$x=-1 x=5$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$16x+20=4x^{2}$$

Subtract 4x^{2} from both sides.

$$16x+20-4x^{2}=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$16x-4x^{2}=-20$$

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.

$$-4x^{2}+16x=-20$$

Divide both sides by -4.

$$\frac{-4x^{2}+16x}{-4}=-\frac{20}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{16}{-4}x=-\frac{20}{-4}$$

Divide 16 by -4.

$$x^{2}-4x=-\frac{20}{-4}$$

Divide -20 by -4.

$$x^{2}-4x=5$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=5+\left(-2\right)^{2}$$

Square -2.

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

Add 5 to 4.

$$x^{2}-4x+4=9$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=3 x-2=-3$$

Add 2 to both sides of the equation.

$$x=5 x=-1$$