Solve 5x+25(x+2)x+25=4x+3

$$5x+25(x+2)x+25=4x+3$$

Solve for x

$$x=\frac{\sqrt{401}-51}{50}\approx -0.619500312$$

$$x=\frac{-\sqrt{401}-51}{50}\approx -1.420499688$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 25 by x+2.

$$5x+\left(25x+50\right)x+25=4x+3$$

Use the distributive property to multiply 25x+50 by x.

$$5x+25x^{2}+50x+25=4x+3$$

Combine 5x and 50x to get 55x.

$$55x+25x^{2}+25=4x+3$$

Subtract 4x from both sides.

$$55x+25x^{2}+25-4x=3$$

Combine 55x and -4x to get 51x.

$$51x+25x^{2}+25=3$$

Subtract 3 from both sides.

$$51x+25x^{2}+25-3=0$$

Subtract 3 from 25 to get 22.

$$51x+25x^{2}+22=0$$

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.

$$25x^{2}+51x+22=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 51 for b, and 22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-51±\sqrt{51^{2}-4\times 25\times 22}}{2\times 25}$$

Square 51.

$$x=\frac{-51±\sqrt{2601-4\times 25\times 22}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-51±\sqrt{2601-100\times 22}}{2\times 25}$$

Multiply -100 times 22.

$$x=\frac{-51±\sqrt{2601-2200}}{2\times 25}$$

Add 2601 to -2200.

$$x=\frac{-51±\sqrt{401}}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-51±\sqrt{401}}{50}$$

Now solve the equation x=\frac{-51±\sqrt{401}}{50} when ± is plus. Add -51 to \sqrt{401}.

$$x=\frac{\sqrt{401}-51}{50}$$

Now solve the equation x=\frac{-51±\sqrt{401}}{50} when ± is minus. Subtract \sqrt{401} from -51.

$$x=\frac{-\sqrt{401}-51}{50}$$

The equation is now solved.

$$x=\frac{\sqrt{401}-51}{50} x=\frac{-\sqrt{401}-51}{50}$$

Steps for Completing the Square

Use the distributive property to multiply 25 by x+2.

$$5x+\left(25x+50\right)x+25=4x+3$$

Use the distributive property to multiply 25x+50 by x.

$$5x+25x^{2}+50x+25=4x+3$$

Combine 5x and 50x to get 55x.

$$55x+25x^{2}+25=4x+3$$

Subtract 4x from both sides.

$$55x+25x^{2}+25-4x=3$$

Combine 55x and -4x to get 51x.

$$51x+25x^{2}+25=3$$

Subtract 25 from both sides.

$$51x+25x^{2}=3-25$$

Subtract 25 from 3 to get -22.

$$51x+25x^{2}=-22$$

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.

$$25x^{2}+51x=-22$$

Divide both sides by 25.

$$\frac{25x^{2}+51x}{25}=-\frac{22}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\frac{51}{25}x=-\frac{22}{25}$$

Divide \frac{51}{25}, the coefficient of the x term, by 2 to get \frac{51}{50}. Then add the square of \frac{51}{50} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{51}{25}x+\left(\frac{51}{50}\right)^{2}=-\frac{22}{25}+\left(\frac{51}{50}\right)^{2}$$

Square \frac{51}{50} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{51}{25}x+\frac{2601}{2500}=-\frac{22}{25}+\frac{2601}{2500}$$

Add -\frac{22}{25} to \frac{2601}{2500} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{51}{25}x+\frac{2601}{2500}=\frac{401}{2500}$$

Factor x^{2}+\frac{51}{25}x+\frac{2601}{2500}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{51}{50}\right)^{2}=\frac{401}{2500}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{51}{50}\right)^{2}}=\sqrt{\frac{401}{2500}}$$

Simplify.

$$x+\frac{51}{50}=\frac{\sqrt{401}}{50} x+\frac{51}{50}=-\frac{\sqrt{401}}{50}$$

Subtract \frac{51}{50} from both sides of the equation.

$$x=\frac{\sqrt{401}-51}{50} x=\frac{-\sqrt{401}-51}{50}$$