Solve 5x^2-2=9x

$$5 x ^ { 2 } - 2 = 9 x$$

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x=2$$

Steps Using Factoring By Grouping

Subtract 9x from both sides.

$$5x^{2}-2-9x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$5x^{2}-9x-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=-9 ab=5\left(-2\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -9.

$$a=-10 b=1$$

Rewrite 5x^{2}-9x-2 as \left(5x^{2}-10x\right)+\left(x-2\right).

$$\left(5x^{2}-10x\right)+\left(x-2\right)$$

Factor out 5x in 5x^{2}-10x.

$$5x\left(x-2\right)+x-2$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(5x+1\right)$$

To find equation solutions, solve x-2=0 and 5x+1=0.

$$x=2 x=-\frac{1}{5}$$

Steps Using the Quadratic Formula

Subtract 9x from both sides.

$$5x^{2}-2-9x=0$$

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.

$$5x^{2}-9x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -9 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 5\left(-2\right)}}{2\times 5}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 5\left(-2\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-9\right)±\sqrt{81-20\left(-2\right)}}{2\times 5}$$

Multiply -20 times -2.

$$x=\frac{-\left(-9\right)±\sqrt{81+40}}{2\times 5}$$

Add 81 to 40.

$$x=\frac{-\left(-9\right)±\sqrt{121}}{2\times 5}$$

Take the square root of 121.

$$x=\frac{-\left(-9\right)±11}{2\times 5}$$

The opposite of -9 is 9.

$$x=\frac{9±11}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{9±11}{10}$$

Now solve the equation x=\frac{9±11}{10} when ± is plus. Add 9 to 11.

$$x=\frac{20}{10}$$

Divide 20 by 10.

$$x=2$$

Now solve the equation x=\frac{9±11}{10} when ± is minus. Subtract 11 from 9.

$$x=-\frac{2}{10}$$

Reduce the fraction \frac{-2}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{5}$$

The equation is now solved.

$$x=2 x=-\frac{1}{5}$$

Steps for Completing the Square

Subtract 9x from both sides.

$$5x^{2}-2-9x=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$5x^{2}-9x=2$$

Divide both sides by 5.

$$\frac{5x^{2}-9x}{5}=\frac{2}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{9}{5}x=\frac{2}{5}$$

Divide -\frac{9}{5}, the coefficient of the x term, by 2 to get -\frac{9}{10}. Then add the square of -\frac{9}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{5}x+\left(-\frac{9}{10}\right)^{2}=\frac{2}{5}+\left(-\frac{9}{10}\right)^{2}$$

Square -\frac{9}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{5}x+\frac{81}{100}=\frac{2}{5}+\frac{81}{100}$$

Add \frac{2}{5} to \frac{81}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{5}x+\frac{81}{100}=\frac{121}{100}$$

Factor x^{2}-\frac{9}{5}x+\frac{81}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{10}\right)^{2}=\frac{121}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{10}\right)^{2}}=\sqrt{\frac{121}{100}}$$

Simplify.

$$x-\frac{9}{10}=\frac{11}{10} x-\frac{9}{10}=-\frac{11}{10}$$

Add \frac{9}{10} to both sides of the equation.

$$x=2 x=-\frac{1}{5}$$