Solve 5k^2=-10k-7

$$5 k ^ { 2 } = - 10 k - 7$$

Solve for k

$$k=\frac{\sqrt{10}i}{5}-1\approx -1+0.632455532i$$

$$k=-\frac{\sqrt{10}i}{5}-1\approx -1-0.632455532i$$

Steps Using the Quadratic Formula

Add 10k to both sides.

$$5k^{2}+10k=-7$$

Add 7 to both sides.

$$5k^{2}+10k+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 10 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-10±\sqrt{10^{2}-4\times 5\times 7}}{2\times 5}$$

Square 10.

$$k=\frac{-10±\sqrt{100-4\times 5\times 7}}{2\times 5}$$

Multiply -4 times 5.

$$k=\frac{-10±\sqrt{100-20\times 7}}{2\times 5}$$

Multiply -20 times 7.

$$k=\frac{-10±\sqrt{100-140}}{2\times 5}$$

Add 100 to -140.

$$k=\frac{-10±\sqrt{-40}}{2\times 5}$$

Take the square root of -40.

$$k=\frac{-10±2\sqrt{10}i}{2\times 5}$$

Multiply 2 times 5.

$$k=\frac{-10±2\sqrt{10}i}{10}$$

Now solve the equation k=\frac{-10±2\sqrt{10}i}{10} when ± is plus. Add -10 to 2i\sqrt{10}.

$$k=\frac{-10+2\sqrt{10}i}{10}$$

Divide -10+2i\sqrt{10} by 10.

$$k=\frac{\sqrt{10}i}{5}-1$$

Now solve the equation k=\frac{-10±2\sqrt{10}i}{10} when ± is minus. Subtract 2i\sqrt{10} from -10.

$$k=\frac{-2\sqrt{10}i-10}{10}$$

Divide -10-2i\sqrt{10} by 10.

$$k=-\frac{\sqrt{10}i}{5}-1$$

The equation is now solved.

$$k=\frac{\sqrt{10}i}{5}-1 k=-\frac{\sqrt{10}i}{5}-1$$

Steps for Completing the Square

Add 10k to both sides.

$$5k^{2}+10k=-7$$

Divide both sides by 5.

$$\frac{5k^{2}+10k}{5}=-\frac{7}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$k^{2}+\frac{10}{5}k=-\frac{7}{5}$$

Divide 10 by 5.

$$k^{2}+2k=-\frac{7}{5}$$

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.

$$k^{2}+2k+1^{2}=-\frac{7}{5}+1^{2}$$

Square 1.

$$k^{2}+2k+1=-\frac{7}{5}+1$$

Add -\frac{7}{5} to 1.

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

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}.

$$\left(k+1\right)^{2}=-\frac{2}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+1\right)^{2}}=\sqrt{-\frac{2}{5}}$$

Simplify.

$$k+1=\frac{\sqrt{10}i}{5} k+1=-\frac{\sqrt{10}i}{5}$$

Subtract 1 from both sides of the equation.

$$k=\frac{\sqrt{10}i}{5}-1 k=-\frac{\sqrt{10}i}{5}-1$$