Solve 5x-4=10x^2

$$5 x - 4 = 10 x ^ { 2 }$$

Solve for x (complex solution)

$$x=-\frac{3\sqrt{15}i}{20}+\frac{1}{4}\approx 0.25-0.580947502i$$

$$x=\frac{3\sqrt{15}i}{20}+\frac{1}{4}\approx 0.25+0.580947502i$$

Steps Using the Quadratic Formula

Subtract 10x^{2} from both sides.

$$5x-4-10x^{2}=0$$

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.

$$-10x^{2}+5x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 5 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-10\right)\left(-4\right)}}{2\left(-10\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-10\right)\left(-4\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-5±\sqrt{25+40\left(-4\right)}}{2\left(-10\right)}$$

Multiply 40 times -4.

$$x=\frac{-5±\sqrt{25-160}}{2\left(-10\right)}$$

Add 25 to -160.

$$x=\frac{-5±\sqrt{-135}}{2\left(-10\right)}$$

Take the square root of -135.

$$x=\frac{-5±3\sqrt{15}i}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-5±3\sqrt{15}i}{-20}$$

Now solve the equation x=\frac{-5±3\sqrt{15}i}{-20} when ± is plus. Add -5 to 3i\sqrt{15}.

$$x=\frac{-5+3\sqrt{15}i}{-20}$$

Divide -5+3i\sqrt{15} by -20.

$$x=-\frac{3\sqrt{15}i}{20}+\frac{1}{4}$$

Now solve the equation x=\frac{-5±3\sqrt{15}i}{-20} when ± is minus. Subtract 3i\sqrt{15} from -5.

$$x=\frac{-3\sqrt{15}i-5}{-20}$$

Divide -5-3i\sqrt{15} by -20.

$$x=\frac{3\sqrt{15}i}{20}+\frac{1}{4}$$

The equation is now solved.

$$x=-\frac{3\sqrt{15}i}{20}+\frac{1}{4} x=\frac{3\sqrt{15}i}{20}+\frac{1}{4}$$

Steps for Completing the Square

Subtract 10x^{2} from both sides.

$$5x-4-10x^{2}=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$5x-10x^{2}=4$$

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.

$$-10x^{2}+5x=4$$

Divide both sides by -10.

$$\frac{-10x^{2}+5x}{-10}=\frac{4}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{5}{-10}x=\frac{4}{-10}$$

Reduce the fraction \frac{5}{-10} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{1}{2}x=\frac{4}{-10}$$

Reduce the fraction \frac{4}{-10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{2}x=-\frac{2}{5}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-\frac{2}{5}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{2}{5}+\frac{1}{16}$$

Add -\frac{2}{5} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{27}{80}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{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}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{27}{80}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{27}{80}}$$

Simplify.

$$x-\frac{1}{4}=\frac{3\sqrt{15}i}{20} x-\frac{1}{4}=-\frac{3\sqrt{15}i}{20}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{3\sqrt{15}i}{20}+\frac{1}{4} x=-\frac{3\sqrt{15}i}{20}+\frac{1}{4}$$