Solve 10x^2=6+9x

$$10 x ^ { 2 } = 6 + 9 x$$

Solve for x

$$x = \frac{\sqrt{321} + 9}{20} \approx 1.345823643$$

$$x=\frac{9-\sqrt{321}}{20}\approx -0.445823643$$

Steps Using the Quadratic Formula

Subtract 6 from both sides.

$$10x^{2}-6=9x$$

Subtract 9x from both sides.

$$10x^{2}-6-9x=0$$

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.

$$10x^{2}-9x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -9 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 10\left(-6\right)}}{2\times 10}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 10\left(-6\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-9\right)±\sqrt{81-40\left(-6\right)}}{2\times 10}$$

Multiply -40 times -6.

$$x=\frac{-\left(-9\right)±\sqrt{81+240}}{2\times 10}$$

Add 81 to 240.

$$x=\frac{-\left(-9\right)±\sqrt{321}}{2\times 10}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{321}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{9±\sqrt{321}}{20}$$

Now solve the equation x=\frac{9±\sqrt{321}}{20} when ± is plus. Add 9 to \sqrt{321}.

$$x=\frac{\sqrt{321}+9}{20}$$

Now solve the equation x=\frac{9±\sqrt{321}}{20} when ± is minus. Subtract \sqrt{321} from 9.

$$x=\frac{9-\sqrt{321}}{20}$$

The equation is now solved.

$$x=\frac{\sqrt{321}+9}{20} x=\frac{9-\sqrt{321}}{20}$$

Steps for Completing the Square

Subtract 9x from both sides.

$$10x^{2}-9x=6$$

Divide both sides by 10.

$$\frac{10x^{2}-9x}{10}=\frac{6}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}-\frac{9}{10}x=\frac{6}{10}$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{9}{10}x=\frac{3}{5}$$

Divide -\frac{9}{10}, the coefficient of the x term, by 2 to get -\frac{9}{20}. Then add the square of -\frac{9}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{10}x+\left(-\frac{9}{20}\right)^{2}=\frac{3}{5}+\left(-\frac{9}{20}\right)^{2}$$

Square -\frac{9}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{10}x+\frac{81}{400}=\frac{3}{5}+\frac{81}{400}$$

Add \frac{3}{5} to \frac{81}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{10}x+\frac{81}{400}=\frac{321}{400}$$

Factor x^{2}-\frac{9}{10}x+\frac{81}{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}.

$$\left(x-\frac{9}{20}\right)^{2}=\frac{321}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{20}\right)^{2}}=\sqrt{\frac{321}{400}}$$

Simplify.

$$x-\frac{9}{20}=\frac{\sqrt{321}}{20} x-\frac{9}{20}=-\frac{\sqrt{321}}{20}$$

Add \frac{9}{20} to both sides of the equation.

$$x=\frac{\sqrt{321}+9}{20} x=\frac{9-\sqrt{321}}{20}$$