Solve x=10x^2+4(x+5)

$$x=10 { x }^{ 2 } +4(x+5)$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{791}i-3}{20}\approx -0.15-1.406236111i$$

$$x=\frac{-3+\sqrt{791}i}{20}\approx -0.15+1.406236111i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4 by x+5.

$$x=10x^{2}+4x+20$$

Subtract 10x^{2} from both sides.

$$x-10x^{2}=4x+20$$

Subtract 4x from both sides.

$$x-10x^{2}-4x=20$$

Combine x and -4x to get -3x.

$$-3x-10x^{2}=20$$

Subtract 20 from both sides.

$$-3x-10x^{2}-20=0$$

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.

$$-10x^{2}-3x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, -3 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-10\right)\left(-20\right)}}{2\left(-10\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-10\right)\left(-20\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-\left(-3\right)±\sqrt{9+40\left(-20\right)}}{2\left(-10\right)}$$

Multiply 40 times -20.

$$x=\frac{-\left(-3\right)±\sqrt{9-800}}{2\left(-10\right)}$$

Add 9 to -800.

$$x=\frac{-\left(-3\right)±\sqrt{-791}}{2\left(-10\right)}$$

Take the square root of -791.

$$x=\frac{-\left(-3\right)±\sqrt{791}i}{2\left(-10\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{791}i}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{3±\sqrt{791}i}{-20}$$

Now solve the equation x=\frac{3±\sqrt{791}i}{-20} when ± is plus. Add 3 to i\sqrt{791}.

$$x=\frac{3+\sqrt{791}i}{-20}$$

Divide 3+i\sqrt{791} by -20.

$$x=\frac{-\sqrt{791}i-3}{20}$$

Now solve the equation x=\frac{3±\sqrt{791}i}{-20} when ± is minus. Subtract i\sqrt{791} from 3.

$$x=\frac{-\sqrt{791}i+3}{-20}$$

Divide 3-i\sqrt{791} by -20.

$$x=\frac{-3+\sqrt{791}i}{20}$$

The equation is now solved.

$$x=\frac{-\sqrt{791}i-3}{20} x=\frac{-3+\sqrt{791}i}{20}$$

Steps for Completing the Square

Use the distributive property to multiply 4 by x+5.

$$x=10x^{2}+4x+20$$

Subtract 10x^{2} from both sides.

$$x-10x^{2}=4x+20$$

Subtract 4x from both sides.

$$x-10x^{2}-4x=20$$

Combine x and -4x to get -3x.

$$-3x-10x^{2}=20$$

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.

$$-10x^{2}-3x=20$$

Divide both sides by -10.

$$\frac{-10x^{2}-3x}{-10}=\frac{20}{-10}$$

Dividing by -10 undoes the multiplication by -10.

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

Divide -3 by -10.

$$x^{2}+\frac{3}{10}x=\frac{20}{-10}$$

Divide 20 by -10.

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

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.

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

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

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

Add -2 to \frac{9}{400}.

$$x^{2}+\frac{3}{10}x+\frac{9}{400}=-\frac{791}{400}$$

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}.

$$\left(x+\frac{3}{20}\right)^{2}=-\frac{791}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{20}\right)^{2}}=\sqrt{-\frac{791}{400}}$$

Simplify.

$$x+\frac{3}{20}=\frac{\sqrt{791}i}{20} x+\frac{3}{20}=-\frac{\sqrt{791}i}{20}$$

Subtract \frac{3}{20} from both sides of the equation.

$$x=\frac{-3+\sqrt{791}i}{20} x=\frac{-\sqrt{791}i-3}{20}$$