Solve x^2-35x+745=0

$${ x }^{ 2 } -35x+745=0$$

Solve for x (complex solution)

$$x=\frac{35+3\sqrt{195}i}{2}\approx 17.5+20.946360066i$$

$$x=\frac{-3\sqrt{195}i+35}{2}\approx 17.5-20.946360066i$$

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.

$$x^{2}-35x+745=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -35 for b, and 745 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 745}}{2}$$

Square -35.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\times 745}}{2}$$

Multiply -4 times 745.

$$x=\frac{-\left(-35\right)±\sqrt{1225-2980}}{2}$$

Add 1225 to -2980.

$$x=\frac{-\left(-35\right)±\sqrt{-1755}}{2}$$

Take the square root of -1755.

$$x=\frac{-\left(-35\right)±3\sqrt{195}i}{2}$$

The opposite of -35 is 35.

$$x=\frac{35±3\sqrt{195}i}{2}$$

Now solve the equation x=\frac{35±3\sqrt{195}i}{2} when ± is plus. Add 35 to 3i\sqrt{195}.

$$x=\frac{35+3\sqrt{195}i}{2}$$

Now solve the equation x=\frac{35±3\sqrt{195}i}{2} when ± is minus. Subtract 3i\sqrt{195} from 35.

$$x=\frac{-3\sqrt{195}i+35}{2}$$

The equation is now solved.

$$x=\frac{35+3\sqrt{195}i}{2} x=\frac{-3\sqrt{195}i+35}{2}$$

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.

$$x^{2}-35x+745=0$$

Subtract 745 from both sides of the equation.

$$x^{2}-35x+745-745=-745$$

Subtracting 745 from itself leaves 0.

$$x^{2}-35x=-745$$

Divide -35, the coefficient of the x term, by 2 to get -\frac{35}{2}. Then add the square of -\frac{35}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-35x+\left(-\frac{35}{2}\right)^{2}=-745+\left(-\frac{35}{2}\right)^{2}$$

Square -\frac{35}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-35x+\frac{1225}{4}=-745+\frac{1225}{4}$$

Add -745 to \frac{1225}{4}.

$$x^{2}-35x+\frac{1225}{4}=-\frac{1755}{4}$$

Factor x^{2}-35x+\frac{1225}{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}.

$$\left(x-\frac{35}{2}\right)^{2}=-\frac{1755}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{2}\right)^{2}}=\sqrt{-\frac{1755}{4}}$$

Simplify.

$$x-\frac{35}{2}=\frac{3\sqrt{195}i}{2} x-\frac{35}{2}=-\frac{3\sqrt{195}i}{2}$$

Add \frac{35}{2} to both sides of the equation.

$$x=\frac{35+3\sqrt{195}i}{2} x=\frac{-3\sqrt{195}i+35}{2}$$