Solve k^2=-2k+35

$$k ^ { 2 } = - 2 k + 35$$

Solve for k

$$k=-7$$

$$k=5$$

Steps Using Factoring

Add 2k to both sides.

$$k^{2}+2k=35$$

Subtract 35 from both sides.

$$k^{2}+2k-35=0$$

To solve the equation, factor k^{2}+2k-35 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+b\right). To find a and b, set up a system to be solved.

$$a+b=2 ab=-35$$

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 -35.

$$-1,35 -5,7$$

Calculate the sum for each pair.

$$-1+35=34 -5+7=2$$

The solution is the pair that gives sum 2.

$$a=-5 b=7$$

Rewrite factored expression \left(k+a\right)\left(k+b\right) using the obtained values.

$$\left(k-5\right)\left(k+7\right)$$

To find equation solutions, solve k-5=0 and k+7=0.

$$k=5 k=-7$$

Steps Using Factoring By Grouping

Add 2k to both sides.

$$k^{2}+2k=35$$

Subtract 35 from both sides.

$$k^{2}+2k-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk-35. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-35\right)=-35$$

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 -35.

$$-1,35 -5,7$$

Calculate the sum for each pair.

$$-1+35=34 -5+7=2$$

The solution is the pair that gives sum 2.

$$a=-5 b=7$$

Rewrite k^{2}+2k-35 as \left(k^{2}-5k\right)+\left(7k-35\right).

$$\left(k^{2}-5k\right)+\left(7k-35\right)$$

Factor out k in the first and 7 in the second group.

$$k\left(k-5\right)+7\left(k-5\right)$$

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

$$\left(k-5\right)\left(k+7\right)$$

To find equation solutions, solve k-5=0 and k+7=0.

$$k=5 k=-7$$

Steps Using the Quadratic Formula

Add 2k to both sides.

$$k^{2}+2k=35$$

Subtract 35 from both sides.

$$k^{2}+2k-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-2±\sqrt{2^{2}-4\left(-35\right)}}{2}$$

Square 2.

$$k=\frac{-2±\sqrt{4-4\left(-35\right)}}{2}$$

Multiply -4 times -35.

$$k=\frac{-2±\sqrt{4+140}}{2}$$

Add 4 to 140.

$$k=\frac{-2±\sqrt{144}}{2}$$

Take the square root of 144.

$$k=\frac{-2±12}{2}$$

Now solve the equation k=\frac{-2±12}{2} when ± is plus. Add -2 to 12.

$$k=\frac{10}{2}$$

Divide 10 by 2.

$$k=5$$

Now solve the equation k=\frac{-2±12}{2} when ± is minus. Subtract 12 from -2.

$$k=-\frac{14}{2}$$

Divide -14 by 2.

$$k=-7$$

The equation is now solved.

$$k=5 k=-7$$

Steps for Completing the Square

Add 2k to both sides.

$$k^{2}+2k=35$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+2k+1^{2}=35+1^{2}$$

Square 1.

$$k^{2}+2k+1=35+1$$

Add 35 to 1.

$$k^{2}+2k+1=36$$

Factor k^{2}+2k+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+1\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+1\right)^{2}}=\sqrt{36}$$

Simplify.

$$k+1=6 k+1=-6$$

Subtract 1 from both sides of the equation.

$$k=5 k=-7$$