Solve 3*(3x+1)-11x=25x^2+1quad3

$$3 \times ( 3 x + 1 ) - 11 x = 25 x ^ { 2 } + 1 \quad 3$$

Solve for x

$$x=-\frac{2}{25}=-0.08$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 3 by 3x+1.

$$9x+3-11x=25x^{2}+1\times 3$$

Combine 9x and -11x to get -2x.

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

Multiply 1 and 3 to get 3.

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

Subtract 25x^{2} from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

$$-2x-25x^{2}=0$$

Factor out x.

$$x\left(-2-25x\right)=0$$

To find equation solutions, solve x=0 and -2-25x=0.

$$x=0 x=-\frac{2}{25}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by 3x+1.

$$9x+3-11x=25x^{2}+1\times 3$$

Combine 9x and -11x to get -2x.

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

Multiply 1 and 3 to get 3.

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

Subtract 25x^{2} from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

$$-2x-25x^{2}=0$$

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.

$$-25x^{2}-2x=0$$

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

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

Take the square root of \left(-2\right)^{2}.

$$x=\frac{-\left(-2\right)±2}{2\left(-25\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{2±2}{-50}$$

Now solve the equation x=\frac{2±2}{-50} when ± is plus. Add 2 to 2.

$$x=\frac{4}{-50}$$

Reduce the fraction \frac{4}{-50} to lowest terms by extracting and canceling out 2.

$$x=-\frac{2}{25}$$

Now solve the equation x=\frac{2±2}{-50} when ± is minus. Subtract 2 from 2.

$$x=\frac{0}{-50}$$

Divide 0 by -50.

$$x=0$$

The equation is now solved.

$$x=-\frac{2}{25} x=0$$

Steps for Completing the Square

Use the distributive property to multiply 3 by 3x+1.

$$9x+3-11x=25x^{2}+1\times 3$$

Combine 9x and -11x to get -2x.

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

Multiply 1 and 3 to get 3.

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

Subtract 25x^{2} from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

$$-2x-25x^{2}=0$$

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.

$$-25x^{2}-2x=0$$

Divide both sides by -25.

$$\frac{-25x^{2}-2x}{-25}=\frac{0}{-25}$$

Dividing by -25 undoes the multiplication by -25.

$$x^{2}+\left(-\frac{2}{-25}\right)x=\frac{0}{-25}$$

Divide -2 by -25.

$$x^{2}+\frac{2}{25}x=\frac{0}{-25}$$

Divide 0 by -25.

$$x^{2}+\frac{2}{25}x=0$$

Divide \frac{2}{25}, the coefficient of the x term, by 2 to get \frac{1}{25}. Then add the square of \frac{1}{25} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{25}x+\left(\frac{1}{25}\right)^{2}=\left(\frac{1}{25}\right)^{2}$$

Square \frac{1}{25} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{25}x+\frac{1}{625}=\frac{1}{625}$$

Factor x^{2}+\frac{2}{25}x+\frac{1}{625}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{25}\right)^{2}=\frac{1}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{25}\right)^{2}}=\sqrt{\frac{1}{625}}$$

Simplify.

$$x+\frac{1}{25}=\frac{1}{25} x+\frac{1}{25}=-\frac{1}{25}$$

Subtract \frac{1}{25} from both sides of the equation.

$$x=0 x=-\frac{2}{25}$$