Solve 112k^2+9k=-7

$$112 k ^ { 2 } + 9 k = - 7$$

Solve for k

$$k=\frac{-9+\sqrt{3055}i}{224}\approx -0.040178571+0.246750243i$$

$$k=\frac{-\sqrt{3055}i-9}{224}\approx -0.040178571-0.246750243i$$

Steps Using the Quadratic Formula

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.

$$112k^{2}+9k=-7$$

Add 7 to both sides of the equation.

$$112k^{2}+9k-\left(-7\right)=-7-\left(-7\right)$$

Subtracting -7 from itself leaves 0.

$$112k^{2}+9k-\left(-7\right)=0$$

Subtract -7 from 0.

$$112k^{2}+9k+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 112 for a, 9 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-9±\sqrt{9^{2}-4\times 112\times 7}}{2\times 112}$$

Square 9.

$$k=\frac{-9±\sqrt{81-4\times 112\times 7}}{2\times 112}$$

Multiply -4 times 112.

$$k=\frac{-9±\sqrt{81-448\times 7}}{2\times 112}$$

Multiply -448 times 7.

$$k=\frac{-9±\sqrt{81-3136}}{2\times 112}$$

Add 81 to -3136.

$$k=\frac{-9±\sqrt{-3055}}{2\times 112}$$

Take the square root of -3055.

$$k=\frac{-9±\sqrt{3055}i}{2\times 112}$$

Multiply 2 times 112.

$$k=\frac{-9±\sqrt{3055}i}{224}$$

Now solve the equation k=\frac{-9±\sqrt{3055}i}{224} when ± is plus. Add -9 to i\sqrt{3055}.

$$k=\frac{-9+\sqrt{3055}i}{224}$$

Now solve the equation k=\frac{-9±\sqrt{3055}i}{224} when ± is minus. Subtract i\sqrt{3055} from -9.

$$k=\frac{-\sqrt{3055}i-9}{224}$$

The equation is now solved.

$$k=\frac{-9+\sqrt{3055}i}{224} k=\frac{-\sqrt{3055}i-9}{224}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$112k^{2}+9k=-7$$

Divide both sides by 112.

$$\frac{112k^{2}+9k}{112}=-\frac{7}{112}$$

Dividing by 112 undoes the multiplication by 112.

$$k^{2}+\frac{9}{112}k=-\frac{7}{112}$$

Reduce the fraction \frac{-7}{112} to lowest terms by extracting and canceling out 7.

$$k^{2}+\frac{9}{112}k=-\frac{1}{16}$$

Divide \frac{9}{112}, the coefficient of the x term, by 2 to get \frac{9}{224}. Then add the square of \frac{9}{224} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+\frac{9}{112}k+\left(\frac{9}{224}\right)^{2}=-\frac{1}{16}+\left(\frac{9}{224}\right)^{2}$$

Square \frac{9}{224} by squaring both the numerator and the denominator of the fraction.

$$k^{2}+\frac{9}{112}k+\frac{81}{50176}=-\frac{1}{16}+\frac{81}{50176}$$

Add -\frac{1}{16} to \frac{81}{50176} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}+\frac{9}{112}k+\frac{81}{50176}=-\frac{3055}{50176}$$

Factor k^{2}+\frac{9}{112}k+\frac{81}{50176}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+\frac{9}{224}\right)^{2}=-\frac{3055}{50176}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{9}{224}\right)^{2}}=\sqrt{-\frac{3055}{50176}}$$

Simplify.

$$k+\frac{9}{224}=\frac{\sqrt{3055}i}{224} k+\frac{9}{224}=-\frac{\sqrt{3055}i}{224}$$

Subtract \frac{9}{224} from both sides of the equation.

$$k=\frac{-9+\sqrt{3055}i}{224} k=\frac{-\sqrt{3055}i-9}{224}$$