Solve 5X(X+2)+6=3

$$5 X ( X + 2 ) + 6 = 3$$

Solve for X

$$X=\frac{\sqrt{10}}{5}-1\approx -0.367544468$$

$$X=-\frac{\sqrt{10}}{5}-1\approx -1.632455532$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5X by X+2.

$$5X^{2}+10X+6=3$$

Subtract 3 from both sides.

$$5X^{2}+10X+6-3=0$$

Subtract 3 from 6 to get 3.

$$5X^{2}+10X+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 10 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$X=\frac{-10±\sqrt{10^{2}-4\times 5\times 3}}{2\times 5}$$

Square 10.

$$X=\frac{-10±\sqrt{100-4\times 5\times 3}}{2\times 5}$$

Multiply -4 times 5.

$$X=\frac{-10±\sqrt{100-20\times 3}}{2\times 5}$$

Multiply -20 times 3.

$$X=\frac{-10±\sqrt{100-60}}{2\times 5}$$

Add 100 to -60.

$$X=\frac{-10±\sqrt{40}}{2\times 5}$$

Take the square root of 40.

$$X=\frac{-10±2\sqrt{10}}{2\times 5}$$

Multiply 2 times 5.

$$X=\frac{-10±2\sqrt{10}}{10}$$

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

$$X=\frac{2\sqrt{10}-10}{10}$$

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

$$X=\frac{\sqrt{10}}{5}-1$$

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

$$X=\frac{-2\sqrt{10}-10}{10}$$

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

$$X=-\frac{\sqrt{10}}{5}-1$$

The equation is now solved.

$$X=\frac{\sqrt{10}}{5}-1 X=-\frac{\sqrt{10}}{5}-1$$

Steps for Completing the Square

Use the distributive property to multiply 5X by X+2.

$$5X^{2}+10X+6=3$$

Subtract 6 from both sides.

$$5X^{2}+10X=3-6$$

Subtract 6 from 3 to get -3.

$$5X^{2}+10X=-3$$

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

Divide 10 by 5.

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

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

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

Square 1.

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

Add -\frac{3}{5} to 1.

$$X^{2}+2X+1=\frac{2}{5}$$

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

$$\left(X+1\right)^{2}=\frac{2}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(X+1\right)^{2}}=\sqrt{\frac{2}{5}}$$

Simplify.

$$X+1=\frac{\sqrt{10}}{5} X+1=-\frac{\sqrt{10}}{5}$$

Subtract 1 from both sides of the equation.

$$X=\frac{\sqrt{10}}{5}-1 X=-\frac{\sqrt{10}}{5}-1$$