Solve x^2+35=5x

$$x ^ { 2 } + 35 = 5 x$$

Solve for x (complex solution)

$$x=\frac{5+\sqrt{115}i}{2}\approx 2.5+5.361902647i$$

$$x=\frac{-\sqrt{115}i+5}{2}\approx 2.5-5.361902647i$$

Steps Using the Quadratic Formula

Subtract 5x from both sides.

$$x^{2}+35-5x=0$$

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.

$$x^{2}-5x+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 35}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 35}}{2}$$

Multiply -4 times 35.

$$x=\frac{-\left(-5\right)±\sqrt{25-140}}{2}$$

Add 25 to -140.

$$x=\frac{-\left(-5\right)±\sqrt{-115}}{2}$$

Take the square root of -115.

$$x=\frac{-\left(-5\right)±\sqrt{115}i}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{115}i}{2}$$

Now solve the equation x=\frac{5±\sqrt{115}i}{2} when ± is plus. Add 5 to i\sqrt{115}.

$$x=\frac{5+\sqrt{115}i}{2}$$

Now solve the equation x=\frac{5±\sqrt{115}i}{2} when ± is minus. Subtract i\sqrt{115} from 5.

$$x=\frac{-\sqrt{115}i+5}{2}$$

The equation is now solved.

$$x=\frac{5+\sqrt{115}i}{2} x=\frac{-\sqrt{115}i+5}{2}$$

Steps for Completing the Square

Subtract 5x from both sides.

$$x^{2}+35-5x=0$$

Subtract 35 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-5x=-35$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-35+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-35+\frac{25}{4}$$

Add -35 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=-\frac{115}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=-\frac{115}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{-\frac{115}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{115}i}{2} x-\frac{5}{2}=-\frac{\sqrt{115}i}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{5+\sqrt{115}i}{2} x=\frac{-\sqrt{115}i+5}{2}$$