Solve 2k^2-5/3=3k

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

Solve for k

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

$$k=5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 3.

$$2k^{2}-5=9k$$

Subtract 9k from both sides.

$$2k^{2}-5-9k=0$$

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

$$2k^{2}-9k-5=0$$

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

$$a+b=-9 ab=2\left(-5\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -9.

$$a=-10 b=1$$

Rewrite 2k^{2}-9k-5 as \left(2k^{2}-10k\right)+\left(k-5\right).

$$\left(2k^{2}-10k\right)+\left(k-5\right)$$

Factor out 2k in 2k^{2}-10k.

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

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

$$\left(k-5\right)\left(2k+1\right)$$

To find equation solutions, solve k-5=0 and 2k+1=0.

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

$$2k^{2}-5=9k$$

Subtract 9k from both sides.

$$2k^{2}-5-9k=0$$

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.

$$2k^{2}-9k-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -9 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

Square -9.

$$k=\frac{-\left(-9\right)±\sqrt{81-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$k=\frac{-\left(-9\right)±\sqrt{81-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$k=\frac{-\left(-9\right)±\sqrt{81+40}}{2\times 2}$$

Add 81 to 40.

$$k=\frac{-\left(-9\right)±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$k=\frac{-\left(-9\right)±11}{2\times 2}$$

The opposite of -9 is 9.

$$k=\frac{9±11}{2\times 2}$$

Multiply 2 times 2.

$$k=\frac{9±11}{4}$$

Now solve the equation k=\frac{9±11}{4} when ± is plus. Add 9 to 11.

$$k=\frac{20}{4}$$

Divide 20 by 4.

$$k=5$$

Now solve the equation k=\frac{9±11}{4} when ± is minus. Subtract 11 from 9.

$$k=-\frac{2}{4}$$

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

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

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 3.

$$2k^{2}-5=9k$$

Subtract 9k from both sides.

$$2k^{2}-5-9k=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$2k^{2}-9k=5$$

Divide both sides by 2.

$$\frac{2k^{2}-9k}{2}=\frac{5}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$k^{2}-\frac{9}{2}k=\frac{5}{2}$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{9}{2}k+\left(-\frac{9}{4}\right)^{2}=\frac{5}{2}+\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{9}{2}k+\frac{81}{16}=\frac{5}{2}+\frac{81}{16}$$

Add \frac{5}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{9}{2}k+\frac{81}{16}=\frac{121}{16}$$

Factor k^{2}-\frac{9}{2}k+\frac{81}{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}.

$$\left(k-\frac{9}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{9}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$k-\frac{9}{4}=\frac{11}{4} k-\frac{9}{4}=-\frac{11}{4}$$

Add \frac{9}{4} to both sides of the equation.

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