Solve 16x+4=20x^2+5x

$$16 x + 4 = 20 x ^ { 2 } + 5 x$$

Solve for x

$$x=\frac{4}{5}=0.8$$

$$x=-\frac{1}{4}=-0.25$$

Steps Using Factoring By Grouping

Subtract 20x^{2} from both sides.

$$16x+4-20x^{2}=5x$$

Subtract 5x from both sides.

$$16x+4-20x^{2}-5x=0$$

Combine 16x and -5x to get 11x.

$$11x+4-20x^{2}=0$$

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

$$-20x^{2}+11x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -20x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=11 ab=-20\times 4=-80$$

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 -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 11.

$$a=16 b=-5$$

Rewrite -20x^{2}+11x+4 as \left(-20x^{2}+16x\right)+\left(-5x+4\right).

$$\left(-20x^{2}+16x\right)+\left(-5x+4\right)$$

Factor out 4x in -20x^{2}+16x.

$$4x\left(-5x+4\right)-5x+4$$

Factor out common term -5x+4 by using distributive property.

$$\left(-5x+4\right)\left(4x+1\right)$$

To find equation solutions, solve -5x+4=0 and 4x+1=0.

$$x=\frac{4}{5} x=-\frac{1}{4}$$

Steps Using the Quadratic Formula

Subtract 20x^{2} from both sides.

$$16x+4-20x^{2}=5x$$

Subtract 5x from both sides.

$$16x+4-20x^{2}-5x=0$$

Combine 16x and -5x to get 11x.

$$11x+4-20x^{2}=0$$

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.

$$-20x^{2}+11x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -20 for a, 11 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-20\right)\times 4}}{2\left(-20\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-20\right)\times 4}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$x=\frac{-11±\sqrt{121+80\times 4}}{2\left(-20\right)}$$

Multiply 80 times 4.

$$x=\frac{-11±\sqrt{121+320}}{2\left(-20\right)}$$

Add 121 to 320.

$$x=\frac{-11±\sqrt{441}}{2\left(-20\right)}$$

Take the square root of 441.

$$x=\frac{-11±21}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{-11±21}{-40}$$

Now solve the equation x=\frac{-11±21}{-40} when ± is plus. Add -11 to 21.

$$x=\frac{10}{-40}$$

Reduce the fraction \frac{10}{-40} to lowest terms by extracting and canceling out 10.

$$x=-\frac{1}{4}$$

Now solve the equation x=\frac{-11±21}{-40} when ± is minus. Subtract 21 from -11.

$$x=-\frac{32}{-40}$$

Reduce the fraction \frac{-32}{-40} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

$$x=-\frac{1}{4} x=\frac{4}{5}$$

Steps for Completing the Square

Subtract 20x^{2} from both sides.

$$16x+4-20x^{2}=5x$$

Subtract 5x from both sides.

$$16x+4-20x^{2}-5x=0$$

Combine 16x and -5x to get 11x.

$$11x+4-20x^{2}=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$11x-20x^{2}=-4$$

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.

$$-20x^{2}+11x=-4$$

Divide both sides by -20.

$$\frac{-20x^{2}+11x}{-20}=-\frac{4}{-20}$$

Dividing by -20 undoes the multiplication by -20.

$$x^{2}+\frac{11}{-20}x=-\frac{4}{-20}$$

Divide 11 by -20.

$$x^{2}-\frac{11}{20}x=-\frac{4}{-20}$$

Reduce the fraction \frac{-4}{-20} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{11}{20}x=\frac{1}{5}$$

Divide -\frac{11}{20}, the coefficient of the x term, by 2 to get -\frac{11}{40}. Then add the square of -\frac{11}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{20}x+\left(-\frac{11}{40}\right)^{2}=\frac{1}{5}+\left(-\frac{11}{40}\right)^{2}$$

Square -\frac{11}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{20}x+\frac{121}{1600}=\frac{1}{5}+\frac{121}{1600}$$

Add \frac{1}{5} to \frac{121}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{20}x+\frac{121}{1600}=\frac{441}{1600}$$

Factor x^{2}-\frac{11}{20}x+\frac{121}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{40}\right)^{2}=\frac{441}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{40}\right)^{2}}=\sqrt{\frac{441}{1600}}$$

Simplify.

$$x-\frac{11}{40}=\frac{21}{40} x-\frac{11}{40}=-\frac{21}{40}$$

Add \frac{11}{40} to both sides of the equation.

$$x=\frac{4}{5} x=-\frac{1}{4}$$