Solve 125x^2-78*5x+36125=0

$$125 x ^ { 2 } - 78 \cdot 5 x + 36125 = 0$$

Solve for x (complex solution)

$$x=\frac{39+4\sqrt{11194}i}{25}\approx 1.56+16.92827221i$$

$$x=\frac{-4\sqrt{11194}i+39}{25}\approx 1.56-16.92827221i$$

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.

$$125x^{2}-390x+36125=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 125 for a, -390 for b, and 36125 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-390\right)±\sqrt{\left(-390\right)^{2}-4\times 125\times 36125}}{2\times 125}$$

Square -390.

$$x=\frac{-\left(-390\right)±\sqrt{152100-4\times 125\times 36125}}{2\times 125}$$

Multiply -4 times 125.

$$x=\frac{-\left(-390\right)±\sqrt{152100-500\times 36125}}{2\times 125}$$

Multiply -500 times 36125.

$$x=\frac{-\left(-390\right)±\sqrt{152100-18062500}}{2\times 125}$$

Add 152100 to -18062500.

$$x=\frac{-\left(-390\right)±\sqrt{-17910400}}{2\times 125}$$

Take the square root of -17910400.

$$x=\frac{-\left(-390\right)±40\sqrt{11194}i}{2\times 125}$$

The opposite of -390 is 390.

$$x=\frac{390±40\sqrt{11194}i}{2\times 125}$$

Multiply 2 times 125.

$$x=\frac{390±40\sqrt{11194}i}{250}$$

Now solve the equation x=\frac{390±40\sqrt{11194}i}{250} when ± is plus. Add 390 to 40i\sqrt{11194}.

$$x=\frac{390+40\sqrt{11194}i}{250}$$

Divide 390+40i\sqrt{11194} by 250.

$$x=\frac{39+4\sqrt{11194}i}{25}$$

Now solve the equation x=\frac{390±40\sqrt{11194}i}{250} when ± is minus. Subtract 40i\sqrt{11194} from 390.

$$x=\frac{-40\sqrt{11194}i+390}{250}$$

Divide 390-40i\sqrt{11194} by 250.

$$x=\frac{-4\sqrt{11194}i+39}{25}$$

The equation is now solved.

$$x=\frac{39+4\sqrt{11194}i}{25} x=\frac{-4\sqrt{11194}i+39}{25}$$

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.

$$125x^{2}-390x+36125=0$$

Subtract 36125 from both sides of the equation.

$$125x^{2}-390x+36125-36125=-36125$$

Subtracting 36125 from itself leaves 0.

$$125x^{2}-390x=-36125$$

Divide both sides by 125.

$$\frac{125x^{2}-390x}{125}=-\frac{36125}{125}$$

Dividing by 125 undoes the multiplication by 125.

$$x^{2}+\left(-\frac{390}{125}\right)x=-\frac{36125}{125}$$

Reduce the fraction \frac{-390}{125} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{78}{25}x=-\frac{36125}{125}$$

Divide -36125 by 125.

$$x^{2}-\frac{78}{25}x=-289$$

Divide -\frac{78}{25}, the coefficient of the x term, by 2 to get -\frac{39}{25}. Then add the square of -\frac{39}{25} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{78}{25}x+\left(-\frac{39}{25}\right)^{2}=-289+\left(-\frac{39}{25}\right)^{2}$$

Square -\frac{39}{25} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{78}{25}x+\frac{1521}{625}=-289+\frac{1521}{625}$$

Add -289 to \frac{1521}{625}.

$$x^{2}-\frac{78}{25}x+\frac{1521}{625}=-\frac{179104}{625}$$

Factor x^{2}-\frac{78}{25}x+\frac{1521}{625}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{39}{25}\right)^{2}=-\frac{179104}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{39}{25}\right)^{2}}=\sqrt{-\frac{179104}{625}}$$

Simplify.

$$x-\frac{39}{25}=\frac{4\sqrt{11194}i}{25} x-\frac{39}{25}=-\frac{4\sqrt{11194}i}{25}$$

Add \frac{39}{25} to both sides of the equation.

$$x=\frac{39+4\sqrt{11194}i}{25} x=\frac{-4\sqrt{11194}i+39}{25}$$