Solve 7x^2+21x+56=14

$$7 { x }^{ 2 } +21x+56 = 14$$

Solve for x (complex solution)

$$x=\frac{-3+\sqrt{15}i}{2}\approx -1.5+1.936491673i$$

$$x=\frac{-\sqrt{15}i-3}{2}\approx -1.5-1.936491673i$$

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.

$$7x^{2}+21x+56=14$$

Subtract 14 from both sides of the equation.

$$7x^{2}+21x+56-14=14-14$$

Subtracting 14 from itself leaves 0.

$$7x^{2}+21x+56-14=0$$

Subtract 14 from 56.

$$7x^{2}+21x+42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 21 for b, and 42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-21±\sqrt{21^{2}-4\times 7\times 42}}{2\times 7}$$

Square 21.

$$x=\frac{-21±\sqrt{441-4\times 7\times 42}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-21±\sqrt{441-28\times 42}}{2\times 7}$$

Multiply -28 times 42.

$$x=\frac{-21±\sqrt{441-1176}}{2\times 7}$$

Add 441 to -1176.

$$x=\frac{-21±\sqrt{-735}}{2\times 7}$$

Take the square root of -735.

$$x=\frac{-21±7\sqrt{15}i}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{-21±7\sqrt{15}i}{14}$$

Now solve the equation x=\frac{-21±7\sqrt{15}i}{14} when ± is plus. Add -21 to 7i\sqrt{15}.

$$x=\frac{-21+7\sqrt{15}i}{14}$$

Divide -21+7i\sqrt{15} by 14.

$$x=\frac{-3+\sqrt{15}i}{2}$$

Now solve the equation x=\frac{-21±7\sqrt{15}i}{14} when ± is minus. Subtract 7i\sqrt{15} from -21.

$$x=\frac{-7\sqrt{15}i-21}{14}$$

Divide -21-7i\sqrt{15} by 14.

$$x=\frac{-\sqrt{15}i-3}{2}$$

The equation is now solved.

$$x=\frac{-3+\sqrt{15}i}{2} x=\frac{-\sqrt{15}i-3}{2}$$

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.

$$7x^{2}+21x+56=14$$

Subtract 56 from both sides of the equation.

$$7x^{2}+21x+56-56=14-56$$

Subtracting 56 from itself leaves 0.

$$7x^{2}+21x=14-56$$

Subtract 56 from 14.

$$7x^{2}+21x=-42$$

Divide both sides by 7.

$$\frac{7x^{2}+21x}{7}=-\frac{42}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}+\frac{21}{7}x=-\frac{42}{7}$$

Divide 21 by 7.

$$x^{2}+3x=-\frac{42}{7}$$

Divide -42 by 7.

$$x^{2}+3x=-6$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-6+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-6+\frac{9}{4}$$

Add -6 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=-\frac{15}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{15}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{15}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{15}i}{2} x+\frac{3}{2}=-\frac{\sqrt{15}i}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{-3+\sqrt{15}i}{2} x=\frac{-\sqrt{15}i-3}{2}$$