Solve 2x-3=5x^2

$$2 x - 3 = 5 x ^ { 2 }$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{14}i+1}{5}\approx 0.2-0.748331477i$$

$$x=\frac{1+\sqrt{14}i}{5}\approx 0.2+0.748331477i$$

Steps Using the Quadratic Formula

Subtract 5x^{2} from both sides.

$$2x-3-5x^{2}=0$$

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.

$$-5x^{2}+2x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-2±\sqrt{4+20\left(-3\right)}}{2\left(-5\right)}$$

Multiply 20 times -3.

$$x=\frac{-2±\sqrt{4-60}}{2\left(-5\right)}$$

Add 4 to -60.

$$x=\frac{-2±\sqrt{-56}}{2\left(-5\right)}$$

Take the square root of -56.

$$x=\frac{-2±2\sqrt{14}i}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-2±2\sqrt{14}i}{-10}$$

Now solve the equation x=\frac{-2±2\sqrt{14}i}{-10} when ± is plus. Add -2 to 2i\sqrt{14}.

$$x=\frac{-2+2\sqrt{14}i}{-10}$$

Divide -2+2i\sqrt{14} by -10.

$$x=\frac{-\sqrt{14}i+1}{5}$$

Now solve the equation x=\frac{-2±2\sqrt{14}i}{-10} when ± is minus. Subtract 2i\sqrt{14} from -2.

$$x=\frac{-2\sqrt{14}i-2}{-10}$$

Divide -2-2i\sqrt{14} by -10.

$$x=\frac{1+\sqrt{14}i}{5}$$

The equation is now solved.

$$x=\frac{-\sqrt{14}i+1}{5} x=\frac{1+\sqrt{14}i}{5}$$

Steps for Completing the Square

Subtract 5x^{2} from both sides.

$$2x-3-5x^{2}=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$2x-5x^{2}=3$$

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.

$$-5x^{2}+2x=3$$

Divide both sides by -5.

$$\frac{-5x^{2}+2x}{-5}=\frac{3}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{2}{-5}x=\frac{3}{-5}$$

Divide 2 by -5.

$$x^{2}-\frac{2}{5}x=\frac{3}{-5}$$

Divide 3 by -5.

$$x^{2}-\frac{2}{5}x=-\frac{3}{5}$$

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=-\frac{3}{5}+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{3}{5}+\frac{1}{25}$$

Add -\frac{3}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{14}{25}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{5}\right)^{2}=-\frac{14}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{-\frac{14}{25}}$$

Simplify.

$$x-\frac{1}{5}=\frac{\sqrt{14}i}{5} x-\frac{1}{5}=-\frac{\sqrt{14}i}{5}$$

Add \frac{1}{5} to both sides of the equation.

$$x=\frac{1+\sqrt{14}i}{5} x=\frac{-\sqrt{14}i+1}{5}$$