Solve 100-24k^2+20k=0

$$100 - 24 k ^ { 2 } + 20 k = 0$$

Solve for k

$$k = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$k = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

Steps Using Factoring By Grouping

Divide both sides by 4.

$$25-6k^{2}+5k=0$$

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

$$-6k^{2}+5k+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -6k^{2}+ak+bk+25. To find a and b, set up a system to be solved.

$$a+b=5 ab=-6\times 25=-150$$

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 -150.

$$-1,150 -2,75 -3,50 -5,30 -6,25 -10,15$$

Calculate the sum for each pair.

$$-1+150=149 -2+75=73 -3+50=47 -5+30=25 -6+25=19 -10+15=5$$

The solution is the pair that gives sum 5.

$$a=15 b=-10$$

Rewrite -6k^{2}+5k+25 as \left(-6k^{2}+15k\right)+\left(-10k+25\right).

$$\left(-6k^{2}+15k\right)+\left(-10k+25\right)$$

Factor out -3k in the first and -5 in the second group.

$$-3k\left(2k-5\right)-5\left(2k-5\right)$$

Factor out common term 2k-5 by using distributive property.

$$\left(2k-5\right)\left(-3k-5\right)$$

To find equation solutions, solve 2k-5=0 and -3k-5=0.

$$k=\frac{5}{2} k=-\frac{5}{3}$$

Steps Using the Quadratic Formula

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.

$$-24k^{2}+20k+100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -24 for a, 20 for b, and 100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-20±\sqrt{20^{2}-4\left(-24\right)\times 100}}{2\left(-24\right)}$$

Square 20.

$$k=\frac{-20±\sqrt{400-4\left(-24\right)\times 100}}{2\left(-24\right)}$$

Multiply -4 times -24.

$$k=\frac{-20±\sqrt{400+96\times 100}}{2\left(-24\right)}$$

Multiply 96 times 100.

$$k=\frac{-20±\sqrt{400+9600}}{2\left(-24\right)}$$

Add 400 to 9600.

$$k=\frac{-20±\sqrt{10000}}{2\left(-24\right)}$$

Take the square root of 10000.

$$k=\frac{-20±100}{2\left(-24\right)}$$

Multiply 2 times -24.

$$k=\frac{-20±100}{-48}$$

Now solve the equation k=\frac{-20±100}{-48} when ± is plus. Add -20 to 100.

$$k=\frac{80}{-48}$$

Reduce the fraction \frac{80}{-48} to lowest terms by extracting and canceling out 16.

$$k=-\frac{5}{3}$$

Now solve the equation k=\frac{-20±100}{-48} when ± is minus. Subtract 100 from -20.

$$k=-\frac{120}{-48}$$

Reduce the fraction \frac{-120}{-48} to lowest terms by extracting and canceling out 24.

$$k=\frac{5}{2}$$

The equation is now solved.

$$k=-\frac{5}{3} k=\frac{5}{2}$$

Steps for Completing the Square

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.

$$-24k^{2}+20k+100=0$$

Subtract 100 from both sides of the equation.

$$-24k^{2}+20k+100-100=-100$$

Subtracting 100 from itself leaves 0.

$$-24k^{2}+20k=-100$$

Divide both sides by -24.

$$\frac{-24k^{2}+20k}{-24}=-\frac{100}{-24}$$

Dividing by -24 undoes the multiplication by -24.

$$k^{2}+\frac{20}{-24}k=-\frac{100}{-24}$$

Reduce the fraction \frac{20}{-24} to lowest terms by extracting and canceling out 4.

$$k^{2}-\frac{5}{6}k=-\frac{100}{-24}$$

Reduce the fraction \frac{-100}{-24} to lowest terms by extracting and canceling out 4.

$$k^{2}-\frac{5}{6}k=\frac{25}{6}$$

Divide -\frac{5}{6}, the coefficient of the x term, by 2 to get -\frac{5}{12}. Then add the square of -\frac{5}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{5}{6}k+\left(-\frac{5}{12}\right)^{2}=\frac{25}{6}+\left(-\frac{5}{12}\right)^{2}$$

Square -\frac{5}{12} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{5}{6}k+\frac{25}{144}=\frac{25}{6}+\frac{25}{144}$$

Add \frac{25}{6} to \frac{25}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{5}{6}k+\frac{25}{144}=\frac{625}{144}$$

Factor k^{2}-\frac{5}{6}k+\frac{25}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{5}{12}\right)^{2}=\frac{625}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{5}{12}\right)^{2}}=\sqrt{\frac{625}{144}}$$

Simplify.

$$k-\frac{5}{12}=\frac{25}{12} k-\frac{5}{12}=-\frac{25}{12}$$

Add \frac{5}{12} to both sides of the equation.

$$k=\frac{5}{2} k=-\frac{5}{3}$$