Solve 5x^2+2=9x+5

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

Solve for x

$$x = \frac{\sqrt{141} + 9}{10} \approx 2.087434209$$

$$x=\frac{9-\sqrt{141}}{10}\approx -0.287434209$$

Steps Using the Quadratic Formula

Subtract 9x from both sides.

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

Subtract 5 from both sides.

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

Subtract 5 from 2 to get -3.

$$5x^{2}-3-9x=0$$

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.

$$5x^{2}-9x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -9 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 5\left(-3\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-9\right)±\sqrt{81-20\left(-3\right)}}{2\times 5}$$

Multiply -20 times -3.

$$x=\frac{-\left(-9\right)±\sqrt{81+60}}{2\times 5}$$

Add 81 to 60.

$$x=\frac{-\left(-9\right)±\sqrt{141}}{2\times 5}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{141}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{9±\sqrt{141}}{10}$$

Now solve the equation x=\frac{9±\sqrt{141}}{10} when ± is plus. Add 9 to \sqrt{141}.

$$x=\frac{\sqrt{141}+9}{10}$$

Now solve the equation x=\frac{9±\sqrt{141}}{10} when ± is minus. Subtract \sqrt{141} from 9.

$$x=\frac{9-\sqrt{141}}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{141}+9}{10} x=\frac{9-\sqrt{141}}{10}$$

Steps for Completing the Square

Subtract 9x from both sides.

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

Subtract 2 from both sides.

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

Subtract 2 from 5 to get 3.

$$5x^{2}-9x=3$$

Divide both sides by 5.

$$\frac{5x^{2}-9x}{5}=\frac{3}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{9}{5}x=\frac{3}{5}$$

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.

$$x^{2}-\frac{9}{5}x+\left(-\frac{9}{10}\right)^{2}=\frac{3}{5}+\left(-\frac{9}{10}\right)^{2}$$

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

$$x^{2}-\frac{9}{5}x+\frac{81}{100}=\frac{3}{5}+\frac{81}{100}$$

Add \frac{3}{5} to \frac{81}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{5}x+\frac{81}{100}=\frac{141}{100}$$

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}.

$$\left(x-\frac{9}{10}\right)^{2}=\frac{141}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{10}\right)^{2}}=\sqrt{\frac{141}{100}}$$

Simplify.

$$x-\frac{9}{10}=\frac{\sqrt{141}}{10} x-\frac{9}{10}=-\frac{\sqrt{141}}{10}$$

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

$$x=\frac{\sqrt{141}+9}{10} x=\frac{9-\sqrt{141}}{10}$$