Solve (x)=3x^2-18x+15/5

$$( x ) = \frac { 3 x ^ { 2 } - 18 x + 15 } { 5 }$$

Solve for x

$$x = \frac{\sqrt{349} + 23}{6} \approx 6.946923615$$

$$x=\frac{23-\sqrt{349}}{6}\approx 0.719743051$$

Steps Using the Quadratic Formula

Divide each term of 3x^{2}-18x+15 by 5 to get \frac{3}{5}x^{2}-\frac{18}{5}x+3.

$$x=\frac{3}{5}x^{2}-\frac{18}{5}x+3$$

Subtract \frac{3}{5}x^{2} from both sides.

$$x-\frac{3}{5}x^{2}=-\frac{18}{5}x+3$$

Add \frac{18}{5}x to both sides.

$$x-\frac{3}{5}x^{2}+\frac{18}{5}x=3$$

Combine x and \frac{18}{5}x to get \frac{23}{5}x.

$$\frac{23}{5}x-\frac{3}{5}x^{2}=3$$

Subtract 3 from both sides.

$$\frac{23}{5}x-\frac{3}{5}x^{2}-3=0$$

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.

$$-\frac{3}{5}x^{2}+\frac{23}{5}x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{5} for a, \frac{23}{5} for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{23}{5}±\sqrt{\left(\frac{23}{5}\right)^{2}-4\left(-\frac{3}{5}\right)\left(-3\right)}}{2\left(-\frac{3}{5}\right)}$$

Square \frac{23}{5} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{23}{5}±\sqrt{\frac{529}{25}-4\left(-\frac{3}{5}\right)\left(-3\right)}}{2\left(-\frac{3}{5}\right)}$$

Multiply -4 times -\frac{3}{5}.

$$x=\frac{-\frac{23}{5}±\sqrt{\frac{529}{25}+\frac{12}{5}\left(-3\right)}}{2\left(-\frac{3}{5}\right)}$$

Multiply \frac{12}{5} times -3.

$$x=\frac{-\frac{23}{5}±\sqrt{\frac{529}{25}-\frac{36}{5}}}{2\left(-\frac{3}{5}\right)}$$

Add \frac{529}{25} to -\frac{36}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{23}{5}±\sqrt{\frac{349}{25}}}{2\left(-\frac{3}{5}\right)}$$

Take the square root of \frac{349}{25}.

$$x=\frac{-\frac{23}{5}±\frac{\sqrt{349}}{5}}{2\left(-\frac{3}{5}\right)}$$

Multiply 2 times -\frac{3}{5}.

$$x=\frac{-\frac{23}{5}±\frac{\sqrt{349}}{5}}{-\frac{6}{5}}$$

Now solve the equation x=\frac{-\frac{23}{5}±\frac{\sqrt{349}}{5}}{-\frac{6}{5}} when ± is plus. Add -\frac{23}{5} to \frac{\sqrt{349}}{5}.

$$x=\frac{\sqrt{349}-23}{-\frac{6}{5}\times 5}$$

Divide \frac{-23+\sqrt{349}}{5} by -\frac{6}{5} by multiplying \frac{-23+\sqrt{349}}{5} by the reciprocal of -\frac{6}{5}.

$$x=\frac{23-\sqrt{349}}{6}$$

Now solve the equation x=\frac{-\frac{23}{5}±\frac{\sqrt{349}}{5}}{-\frac{6}{5}} when ± is minus. Subtract \frac{\sqrt{349}}{5} from -\frac{23}{5}.

$$x=\frac{-\sqrt{349}-23}{-\frac{6}{5}\times 5}$$

Divide \frac{-23-\sqrt{349}}{5} by -\frac{6}{5} by multiplying \frac{-23-\sqrt{349}}{5} by the reciprocal of -\frac{6}{5}.

$$x=\frac{\sqrt{349}+23}{6}$$

The equation is now solved.

$$x=\frac{23-\sqrt{349}}{6} x=\frac{\sqrt{349}+23}{6}$$

Steps for Completing the Square

Divide each term of 3x^{2}-18x+15 by 5 to get \frac{3}{5}x^{2}-\frac{18}{5}x+3.

$$x=\frac{3}{5}x^{2}-\frac{18}{5}x+3$$

Subtract \frac{3}{5}x^{2} from both sides.

$$x-\frac{3}{5}x^{2}=-\frac{18}{5}x+3$$

Add \frac{18}{5}x to both sides.

$$x-\frac{3}{5}x^{2}+\frac{18}{5}x=3$$

Combine x and \frac{18}{5}x to get \frac{23}{5}x.

$$\frac{23}{5}x-\frac{3}{5}x^{2}=3$$

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.

$$-\frac{3}{5}x^{2}+\frac{23}{5}x=3$$

Divide both sides of the equation by -\frac{3}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{5}x^{2}+\frac{23}{5}x}{-\frac{3}{5}}=\frac{3}{-\frac{3}{5}}$$

Dividing by -\frac{3}{5} undoes the multiplication by -\frac{3}{5}.

$$x^{2}+\frac{\frac{23}{5}}{-\frac{3}{5}}x=\frac{3}{-\frac{3}{5}}$$

Divide \frac{23}{5} by -\frac{3}{5} by multiplying \frac{23}{5} by the reciprocal of -\frac{3}{5}.

$$x^{2}-\frac{23}{3}x=\frac{3}{-\frac{3}{5}}$$

Divide 3 by -\frac{3}{5} by multiplying 3 by the reciprocal of -\frac{3}{5}.

$$x^{2}-\frac{23}{3}x=-5$$

Divide -\frac{23}{3}, the coefficient of the x term, by 2 to get -\frac{23}{6}. Then add the square of -\frac{23}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{23}{3}x+\left(-\frac{23}{6}\right)^{2}=-5+\left(-\frac{23}{6}\right)^{2}$$

Square -\frac{23}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{23}{3}x+\frac{529}{36}=-5+\frac{529}{36}$$

Add -5 to \frac{529}{36}.

$$x^{2}-\frac{23}{3}x+\frac{529}{36}=\frac{349}{36}$$

Factor x^{2}-\frac{23}{3}x+\frac{529}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{23}{6}\right)^{2}=\frac{349}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{23}{6}\right)^{2}}=\sqrt{\frac{349}{36}}$$

Simplify.

$$x-\frac{23}{6}=\frac{\sqrt{349}}{6} x-\frac{23}{6}=-\frac{\sqrt{349}}{6}$$

Add \frac{23}{6} to both sides of the equation.

$$x=\frac{\sqrt{349}+23}{6} x=\frac{23-\sqrt{349}}{6}$$