Solve 3x^2+1-7x=13x^2

$$3 x ^ { 2 } + 1 - 7 x = 13 x ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{89}-7}{20}\approx 0.121699057$$

$$x=\frac{-\sqrt{89}-7}{20}\approx -0.821699057$$

Steps Using the Quadratic Formula

Subtract 13x^{2} from both sides.

$$3x^{2}+1-7x-13x^{2}=0$$

Combine 3x^{2} and -13x^{2} to get -10x^{2}.

$$-10x^{2}+1-7x=0$$

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.

$$-10x^{2}-7x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, -7 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-10\right)}}{2\left(-10\right)}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-10\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-\left(-7\right)±\sqrt{49+40}}{2\left(-10\right)}$$

Add 49 to 40.

$$x=\frac{-\left(-7\right)±\sqrt{89}}{2\left(-10\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{89}}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{7±\sqrt{89}}{-20}$$

Now solve the equation x=\frac{7±\sqrt{89}}{-20} when ± is plus. Add 7 to \sqrt{89}.

$$x=\frac{\sqrt{89}+7}{-20}$$

Divide 7+\sqrt{89} by -20.

$$x=\frac{-\sqrt{89}-7}{20}$$

Now solve the equation x=\frac{7±\sqrt{89}}{-20} when ± is minus. Subtract \sqrt{89} from 7.

$$x=\frac{7-\sqrt{89}}{-20}$$

Divide 7-\sqrt{89} by -20.

$$x=\frac{\sqrt{89}-7}{20}$$

The equation is now solved.

$$x=\frac{-\sqrt{89}-7}{20} x=\frac{\sqrt{89}-7}{20}$$

Steps for Completing the Square

Subtract 13x^{2} from both sides.

$$3x^{2}+1-7x-13x^{2}=0$$

Combine 3x^{2} and -13x^{2} to get -10x^{2}.

$$-10x^{2}+1-7x=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-10x^{2}-7x=-1$$

Divide both sides by -10.

$$\frac{-10x^{2}-7x}{-10}=-\frac{1}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\left(-\frac{7}{-10}\right)x=-\frac{1}{-10}$$

Divide -7 by -10.

$$x^{2}+\frac{7}{10}x=-\frac{1}{-10}$$

Divide -1 by -10.

$$x^{2}+\frac{7}{10}x=\frac{1}{10}$$

Divide \frac{7}{10}, the coefficient of the x term, by 2 to get \frac{7}{20}. Then add the square of \frac{7}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{10}x+\left(\frac{7}{20}\right)^{2}=\frac{1}{10}+\left(\frac{7}{20}\right)^{2}$$

Square \frac{7}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{10}x+\frac{49}{400}=\frac{1}{10}+\frac{49}{400}$$

Add \frac{1}{10} to \frac{49}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{10}x+\frac{49}{400}=\frac{89}{400}$$

Factor x^{2}+\frac{7}{10}x+\frac{49}{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}.

$$\left(x+\frac{7}{20}\right)^{2}=\frac{89}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{20}\right)^{2}}=\sqrt{\frac{89}{400}}$$

Simplify.

$$x+\frac{7}{20}=\frac{\sqrt{89}}{20} x+\frac{7}{20}=-\frac{\sqrt{89}}{20}$$

Subtract \frac{7}{20} from both sides of the equation.

$$x=\frac{\sqrt{89}-7}{20} x=\frac{-\sqrt{89}-7}{20}$$