Solve x^2-49=9-{left(x-7/2right)}^2

$${ x }^{ 2 } -49=9- { \left(x- \frac{ 7 }{ 2 } \right) }^{ 2 }$$

Solve for x

$$x = \frac{\sqrt{415} + 7}{4} \approx 6.842887197$$

$$x=\frac{7-\sqrt{415}}{4}\approx -3.342887197$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{7}{2}\right)^{2}.

$$x^{2}-49=9-\left(x^{2}-7x+\frac{49}{4}\right)$$

To find the opposite of x^{2}-7x+\frac{49}{4}, find the opposite of each term.

$$x^{2}-49=9-x^{2}+7x-\frac{49}{4}$$

Subtract \frac{49}{4} from 9 to get -\frac{13}{4}.

$$x^{2}-49=-\frac{13}{4}-x^{2}+7x$$

Subtract -\frac{13}{4} from both sides.

$$x^{2}-49-\left(-\frac{13}{4}\right)=-x^{2}+7x$$

The opposite of -\frac{13}{4} is \frac{13}{4}.

$$x^{2}-49+\frac{13}{4}=-x^{2}+7x$$

Add x^{2} to both sides.

$$x^{2}-49+\frac{13}{4}+x^{2}=7x$$

Add -49 and \frac{13}{4} to get -\frac{183}{4}.

$$x^{2}-\frac{183}{4}+x^{2}=7x$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-\frac{183}{4}=7x$$

Subtract 7x from both sides.

$$2x^{2}-\frac{183}{4}-7x=0$$

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.

$$2x^{2}-7x-\frac{183}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and -\frac{183}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-\frac{183}{4}\right)}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\left(-\frac{183}{4}\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\left(-\frac{183}{4}\right)}}{2\times 2}$$

Multiply -8 times -\frac{183}{4}.

$$x=\frac{-\left(-7\right)±\sqrt{49+366}}{2\times 2}$$

Add 49 to 366.

$$x=\frac{-\left(-7\right)±\sqrt{415}}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{415}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±\sqrt{415}}{4}$$

Now solve the equation x=\frac{7±\sqrt{415}}{4} when ± is plus. Add 7 to \sqrt{415}.

$$x=\frac{\sqrt{415}+7}{4}$$

Now solve the equation x=\frac{7±\sqrt{415}}{4} when ± is minus. Subtract \sqrt{415} from 7.

$$x=\frac{7-\sqrt{415}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{415}+7}{4} x=\frac{7-\sqrt{415}}{4}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{7}{2}\right)^{2}.

$$x^{2}-49=9-\left(x^{2}-7x+\frac{49}{4}\right)$$

To find the opposite of x^{2}-7x+\frac{49}{4}, find the opposite of each term.

$$x^{2}-49=9-x^{2}+7x-\frac{49}{4}$$

Subtract \frac{49}{4} from 9 to get -\frac{13}{4}.

$$x^{2}-49=-\frac{13}{4}-x^{2}+7x$$

Add x^{2} to both sides.

$$x^{2}-49+x^{2}=-\frac{13}{4}+7x$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-49=-\frac{13}{4}+7x$$

Subtract 7x from both sides.

$$2x^{2}-49-7x=-\frac{13}{4}$$

Add 49 to both sides.

$$2x^{2}-7x=-\frac{13}{4}+49$$

Add -\frac{13}{4} and 49 to get \frac{183}{4}.

$$2x^{2}-7x=\frac{183}{4}$$

Divide both sides by 2.

$$\frac{2x^{2}-7x}{2}=\frac{\frac{183}{4}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{7}{2}x=\frac{\frac{183}{4}}{2}$$

Divide \frac{183}{4} by 2.

$$x^{2}-\frac{7}{2}x=\frac{183}{8}$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=\frac{183}{8}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{183}{8}+\frac{49}{16}$$

Add \frac{183}{8} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{415}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{415}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{415}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{\sqrt{415}}{4} x-\frac{7}{4}=-\frac{\sqrt{415}}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{\sqrt{415}+7}{4} x=\frac{7-\sqrt{415}}{4}$$