Solve 20k^2+5=29k

$$20 k ^ { 2 } + 5 = 29 k$$

Solve for k

$$k=\frac{1}{5}=0.2$$

$$k = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps Using Factoring By Grouping

Subtract 29k from both sides.

$$20k^{2}+5-29k=0$$

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

$$20k^{2}-29k+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 20k^{2}+ak+bk+5. To find a and b, set up a system to be solved.

$$a+b=-29 ab=20\times 5=100$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -29.

$$a=-25 b=-4$$

Rewrite 20k^{2}-29k+5 as \left(20k^{2}-25k\right)+\left(-4k+5\right).

$$\left(20k^{2}-25k\right)+\left(-4k+5\right)$$

Factor out 5k in the first and -1 in the second group.

$$5k\left(4k-5\right)-\left(4k-5\right)$$

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

$$\left(4k-5\right)\left(5k-1\right)$$

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

$$k=\frac{5}{4} k=\frac{1}{5}$$

Steps Using the Quadratic Formula

Subtract 29k from both sides.

$$20k^{2}+5-29k=0$$

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.

$$20k^{2}-29k+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -29 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 20\times 5}}{2\times 20}$$

Square -29.

$$k=\frac{-\left(-29\right)±\sqrt{841-4\times 20\times 5}}{2\times 20}$$

Multiply -4 times 20.

$$k=\frac{-\left(-29\right)±\sqrt{841-80\times 5}}{2\times 20}$$

Multiply -80 times 5.

$$k=\frac{-\left(-29\right)±\sqrt{841-400}}{2\times 20}$$

Add 841 to -400.

$$k=\frac{-\left(-29\right)±\sqrt{441}}{2\times 20}$$

Take the square root of 441.

$$k=\frac{-\left(-29\right)±21}{2\times 20}$$

The opposite of -29 is 29.

$$k=\frac{29±21}{2\times 20}$$

Multiply 2 times 20.

$$k=\frac{29±21}{40}$$

Now solve the equation k=\frac{29±21}{40} when ± is plus. Add 29 to 21.

$$k=\frac{50}{40}$$

Reduce the fraction \frac{50}{40} to lowest terms by extracting and canceling out 10.

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

Now solve the equation k=\frac{29±21}{40} when ± is minus. Subtract 21 from 29.

$$k=\frac{8}{40}$$

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

$$k=\frac{1}{5}$$

The equation is now solved.

$$k=\frac{5}{4} k=\frac{1}{5}$$

Steps for Completing the Square

Subtract 29k from both sides.

$$20k^{2}+5-29k=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$20k^{2}-29k=-5$$

Divide both sides by 20.

$$\frac{20k^{2}-29k}{20}=-\frac{5}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$k^{2}-\frac{29}{20}k=-\frac{5}{20}$$

Reduce the fraction \frac{-5}{20} to lowest terms by extracting and canceling out 5.

$$k^{2}-\frac{29}{20}k=-\frac{1}{4}$$

Divide -\frac{29}{20}, the coefficient of the x term, by 2 to get -\frac{29}{40}. Then add the square of -\frac{29}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{29}{20}k+\left(-\frac{29}{40}\right)^{2}=-\frac{1}{4}+\left(-\frac{29}{40}\right)^{2}$$

Square -\frac{29}{40} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{29}{20}k+\frac{841}{1600}=-\frac{1}{4}+\frac{841}{1600}$$

Add -\frac{1}{4} to \frac{841}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{29}{20}k+\frac{841}{1600}=\frac{441}{1600}$$

Factor k^{2}-\frac{29}{20}k+\frac{841}{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}.

$$\left(k-\frac{29}{40}\right)^{2}=\frac{441}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{29}{40}\right)^{2}}=\sqrt{\frac{441}{1600}}$$

Simplify.

$$k-\frac{29}{40}=\frac{21}{40} k-\frac{29}{40}=-\frac{21}{40}$$

Add \frac{29}{40} to both sides of the equation.

$$k=\frac{5}{4} k=\frac{1}{5}$$