Solve 9x^2+169=78x

$$9 x ^ { 2 } + 169 = 78 x$$

Solve for x

$$x = \frac{13}{3} = 4\frac{1}{3} \approx 4.333333333$$

Steps Using the Quadratic Formula

Subtract 78x from both sides.

$$9x^{2}+169-78x=0$$

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.

$$9x^{2}-78x+169=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -78 for b, and 169 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-78\right)±\sqrt{\left(-78\right)^{2}-4\times 9\times 169}}{2\times 9}$$

Square -78.

$$x=\frac{-\left(-78\right)±\sqrt{6084-4\times 9\times 169}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-78\right)±\sqrt{6084-36\times 169}}{2\times 9}$$

Multiply -36 times 169.

$$x=\frac{-\left(-78\right)±\sqrt{6084-6084}}{2\times 9}$$

Add 6084 to -6084.

$$x=\frac{-\left(-78\right)±\sqrt{0}}{2\times 9}$$

Take the square root of 0.

$$x=-\frac{-78}{2\times 9}$$

The opposite of -78 is 78.

$$x=\frac{78}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{78}{18}$$

Reduce the fraction \frac{78}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{13}{3}$$

Steps for Completing the Square

Subtract 78x from both sides.

$$9x^{2}+169-78x=0$$

Subtract 169 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-78x=-169$$

Divide both sides by 9.

$$\frac{9x^{2}-78x}{9}=-\frac{169}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{78}{9}\right)x=-\frac{169}{9}$$

Reduce the fraction \frac{-78}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{26}{3}x=-\frac{169}{9}$$

Divide -\frac{26}{3}, the coefficient of the x term, by 2 to get -\frac{13}{3}. Then add the square of -\frac{13}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{26}{3}x+\left(-\frac{13}{3}\right)^{2}=-\frac{169}{9}+\left(-\frac{13}{3}\right)^{2}$$

Square -\frac{13}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{26}{3}x+\frac{169}{9}=\frac{-169+169}{9}$$

Add -\frac{169}{9} to \frac{169}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{26}{3}x+\frac{169}{9}=0$$

Factor x^{2}-\frac{26}{3}x+\frac{169}{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}.

$$\left(x-\frac{13}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{13}{3}=0 x-\frac{13}{3}=0$$

Add \frac{13}{3} to both sides of the equation.

$$x=\frac{13}{3} x=\frac{13}{3}$$

The equation is now solved. Solutions are the same.

$$x=\frac{13}{3}$$