Solve 3/8=4x^2+16-20/2*2x*4

$$\frac { 3 } { 8 } = \frac { 4 x ^ { 2 } + 16 - 20 } { 2 \times 2 x \times 4 }$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 16x, the least common multiple of 8,2\times 2x\times 4.

$$6x=4x^{2}+16-20$$

Subtract 20 from 16 to get -4.

$$6x=4x^{2}-4$$

Subtract 4x^{2} from both sides.

$$6x-4x^{2}=-4$$

Add 4 to both sides.

$$6x-4x^{2}+4=0$$

Divide both sides by 2.

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

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

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

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

$$a+b=3 ab=-2\times 2=-4$$

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. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=4 b=-1$$

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

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

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

$$2x\left(-x+2\right)-x+2$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(2x+1\right)$$

To find equation solutions, solve -x+2=0 and 2x+1=0.

$$x=2 x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 16x, the least common multiple of 8,2\times 2x\times 4.

$$6x=4x^{2}+16-20$$

Subtract 20 from 16 to get -4.

$$6x=4x^{2}-4$$

Subtract 4x^{2} from both sides.

$$6x-4x^{2}=-4$$

Add 4 to both sides.

$$6x-4x^{2}+4=0$$

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.

$$-4x^{2}+6x+4=0$$

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

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

Square 6.

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

Multiply -4 times -4.

$$x=\frac{-6±\sqrt{36+16\times 4}}{2\left(-4\right)}$$

Multiply 16 times 4.

$$x=\frac{-6±\sqrt{36+64}}{2\left(-4\right)}$$

Add 36 to 64.

$$x=\frac{-6±\sqrt{100}}{2\left(-4\right)}$$

Take the square root of 100.

$$x=\frac{-6±10}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-6±10}{-8}$$

Now solve the equation x=\frac{-6±10}{-8} when ± is plus. Add -6 to 10.

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

Reduce the fraction \frac{4}{-8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{1}{2}$$

Now solve the equation x=\frac{-6±10}{-8} when ± is minus. Subtract 10 from -6.

$$x=-\frac{16}{-8}$$

Divide -16 by -8.

$$x=2$$

The equation is now solved.

$$x=-\frac{1}{2} x=2$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 16x, the least common multiple of 8,2\times 2x\times 4.

$$6x=4x^{2}+16-20$$

Subtract 20 from 16 to get -4.

$$6x=4x^{2}-4$$

Subtract 4x^{2} from both sides.

$$6x-4x^{2}=-4$$

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.

$$-4x^{2}+6x=-4$$

Divide both sides by -4.

$$\frac{-4x^{2}+6x}{-4}=-\frac{4}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{6}{-4}x=-\frac{4}{-4}$$

Reduce the fraction \frac{6}{-4} to lowest terms by extracting and canceling out 2.

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

Divide -4 by -4.

$$x^{2}-\frac{3}{2}x=1$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=1+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=1+\frac{9}{16}$$

Add 1 to \frac{9}{16}.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{5}{4} x-\frac{3}{4}=-\frac{5}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=2 x=-\frac{1}{2}$$