Solve k(2k+5)-3=0

$$k ( 2 k + 5 ) - 3 = 0$$

Solve for k

$$k=-3$$

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

Steps Using Factoring By Grouping

Use the distributive property to multiply k by 2k+5.

$$2k^{2}+5k-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2k^{2}+ak+bk-3. To find a and b, set up a system to be solved.

$$a+b=5 ab=2\left(-3\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 5.

$$a=-1 b=6$$

Rewrite 2k^{2}+5k-3 as \left(2k^{2}-k\right)+\left(6k-3\right).

$$\left(2k^{2}-k\right)+\left(6k-3\right)$$

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

$$k\left(2k-1\right)+3\left(2k-1\right)$$

Factor out common term 2k-1 by using distributive property.

$$\left(2k-1\right)\left(k+3\right)$$

To find equation solutions, solve 2k-1=0 and k+3=0.

$$k=\frac{1}{2} k=-3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply k by 2k+5.

$$2k^{2}+5k-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-5±\sqrt{5^{2}-4\times 2\left(-3\right)}}{2\times 2}$$

Square 5.

$$k=\frac{-5±\sqrt{25-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$k=\frac{-5±\sqrt{25-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$k=\frac{-5±\sqrt{25+24}}{2\times 2}$$

Add 25 to 24.

$$k=\frac{-5±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$k=\frac{-5±7}{2\times 2}$$

Multiply 2 times 2.

$$k=\frac{-5±7}{4}$$

Now solve the equation k=\frac{-5±7}{4} when ± is plus. Add -5 to 7.

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

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

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

Now solve the equation k=\frac{-5±7}{4} when ± is minus. Subtract 7 from -5.

$$k=-\frac{12}{4}$$

Divide -12 by 4.

$$k=-3$$

The equation is now solved.

$$k=\frac{1}{2} k=-3$$

Steps for Completing the Square

Use the distributive property to multiply k by 2k+5.

$$2k^{2}+5k-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$2k^{2}+5k=3$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+\frac{5}{2}k+\left(\frac{5}{4}\right)^{2}=\frac{3}{2}+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$k^{2}+\frac{5}{2}k+\frac{25}{16}=\frac{3}{2}+\frac{25}{16}$$

Add \frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}+\frac{5}{2}k+\frac{25}{16}=\frac{49}{16}$$

Factor k^{2}+\frac{5}{2}k+\frac{25}{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}.

$$\left(k+\frac{5}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{5}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$k+\frac{5}{4}=\frac{7}{4} k+\frac{5}{4}=-\frac{7}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$k=\frac{1}{2} k=-3$$