Solve 3x/2=5x^2+3

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

Solve for x (complex solution)

$$x=\frac{3+\sqrt{231}i}{20}\approx 0.15+0.759934208i$$

$$x=\frac{-\sqrt{231}i+3}{20}\approx 0.15-0.759934208i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$3x=10x^{2}+6$$

Subtract 10x^{2} from both sides.

$$3x-10x^{2}=6$$

Subtract 6 from both sides.

$$3x-10x^{2}-6=0$$

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.

$$-10x^{2}+3x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 3 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-10\right)\left(-6\right)}}{2\left(-10\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-10\right)\left(-6\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-3±\sqrt{9+40\left(-6\right)}}{2\left(-10\right)}$$

Multiply 40 times -6.

$$x=\frac{-3±\sqrt{9-240}}{2\left(-10\right)}$$

Add 9 to -240.

$$x=\frac{-3±\sqrt{-231}}{2\left(-10\right)}$$

Take the square root of -231.

$$x=\frac{-3±\sqrt{231}i}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-3±\sqrt{231}i}{-20}$$

Now solve the equation x=\frac{-3±\sqrt{231}i}{-20} when ± is plus. Add -3 to i\sqrt{231}.

$$x=\frac{-3+\sqrt{231}i}{-20}$$

Divide -3+i\sqrt{231} by -20.

$$x=\frac{-\sqrt{231}i+3}{20}$$

Now solve the equation x=\frac{-3±\sqrt{231}i}{-20} when ± is minus. Subtract i\sqrt{231} from -3.

$$x=\frac{-\sqrt{231}i-3}{-20}$$

Divide -3-i\sqrt{231} by -20.

$$x=\frac{3+\sqrt{231}i}{20}$$

The equation is now solved.

$$x=\frac{-\sqrt{231}i+3}{20} x=\frac{3+\sqrt{231}i}{20}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$3x=10x^{2}+6$$

Subtract 10x^{2} from both sides.

$$3x-10x^{2}=6$$

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.

$$-10x^{2}+3x=6$$

Divide both sides by -10.

$$\frac{-10x^{2}+3x}{-10}=\frac{6}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{3}{-10}x=\frac{6}{-10}$$

Divide 3 by -10.

$$x^{2}-\frac{3}{10}x=\frac{6}{-10}$$

Reduce the fraction \frac{6}{-10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{10}x=-\frac{3}{5}$$

Divide -\frac{3}{10}, the coefficient of the x term, by 2 to get -\frac{3}{20}. Then add the square of -\frac{3}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{10}x+\left(-\frac{3}{20}\right)^{2}=-\frac{3}{5}+\left(-\frac{3}{20}\right)^{2}$$

Square -\frac{3}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{10}x+\frac{9}{400}=-\frac{3}{5}+\frac{9}{400}$$

Add -\frac{3}{5} to \frac{9}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{10}x+\frac{9}{400}=-\frac{231}{400}$$

Factor x^{2}-\frac{3}{10}x+\frac{9}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{20}\right)^{2}=-\frac{231}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{20}\right)^{2}}=\sqrt{-\frac{231}{400}}$$

Simplify.

$$x-\frac{3}{20}=\frac{\sqrt{231}i}{20} x-\frac{3}{20}=-\frac{\sqrt{231}i}{20}$$

Add \frac{3}{20} to both sides of the equation.

$$x=\frac{3+\sqrt{231}i}{20} x=\frac{-\sqrt{231}i+3}{20}$$