Solve 2k^2=-10+12k

$$2 k ^ { 2 } = - 10 + 12 k$$

Solve for k

$$k=1$$

$$k=5$$

Steps Using Factoring By Grouping

Subtract -10 from both sides.

$$2k^{2}-\left(-10\right)=12k$$

The opposite of -10 is 10.

$$2k^{2}+10=12k$$

Subtract 12k from both sides.

$$2k^{2}+10-12k=0$$

Divide both sides by 2.

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

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

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

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

$$a+b=-6 ab=1\times 5=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

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

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

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

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

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

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

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

$$k=5 k=1$$

Steps Using the Quadratic Formula

Subtract -10 from both sides.

$$2k^{2}-\left(-10\right)=12k$$

The opposite of -10 is 10.

$$2k^{2}+10=12k$$

Subtract 12k from both sides.

$$2k^{2}+10-12k=0$$

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.

$$2k^{2}-12k+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -12 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 10}}{2\times 2}$$

Square -12.

$$k=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 10}}{2\times 2}$$

Multiply -4 times 2.

$$k=\frac{-\left(-12\right)±\sqrt{144-8\times 10}}{2\times 2}$$

Multiply -8 times 10.

$$k=\frac{-\left(-12\right)±\sqrt{144-80}}{2\times 2}$$

Add 144 to -80.

$$k=\frac{-\left(-12\right)±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$k=\frac{-\left(-12\right)±8}{2\times 2}$$

The opposite of -12 is 12.

$$k=\frac{12±8}{2\times 2}$$

Multiply 2 times 2.

$$k=\frac{12±8}{4}$$

Now solve the equation k=\frac{12±8}{4} when ± is plus. Add 12 to 8.

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

Divide 20 by 4.

$$k=5$$

Now solve the equation k=\frac{12±8}{4} when ± is minus. Subtract 8 from 12.

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

Divide 4 by 4.

$$k=1$$

The equation is now solved.

$$k=5 k=1$$

Steps for Completing the Square

Subtract 12k from both sides.

$$2k^{2}-12k=-10$$

Divide both sides by 2.

$$\frac{2k^{2}-12k}{2}=-\frac{10}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$k^{2}+\left(-\frac{12}{2}\right)k=-\frac{10}{2}$$

Divide -12 by 2.

$$k^{2}-6k=-\frac{10}{2}$$

Divide -10 by 2.

$$k^{2}-6k=-5$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-6k+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$k^{2}-6k+9=-5+9$$

Add -5 to 9.

$$k^{2}-6k+9=4$$

Factor k^{2}-6k+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-3\right)^{2}=4$$

Take the square root of both sides of the equation.

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

Simplify.

$$k-3=2 k-3=-2$$

Add 3 to both sides of the equation.

$$k=5 k=1$$