Solve 9div6/5=x^2-2x+3x

$$9 \div \frac{ 6 }{ 5 } = { x }^{ 2 } -2x+3x$$

Solve for x

$$x = \frac{\sqrt{31} - 1}{2} \approx 2.283882181$$

$$x=\frac{-\sqrt{31}-1}{2}\approx -3.283882181$$

Steps Using the Quadratic Formula

Divide 9 by \frac{6}{5} by multiplying 9 by the reciprocal of \frac{6}{5}.

$$9\times \frac{5}{6}=x^{2}-2x+3x$$

Multiply 9 and \frac{5}{6} to get \frac{15}{2}.

$$\frac{15}{2}=x^{2}-2x+3x$$

Combine -2x and 3x to get x.

$$\frac{15}{2}=x^{2}+x$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+x=\frac{15}{2}$$

Subtract \frac{15}{2} from both sides.

$$x^{2}+x-\frac{15}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -\frac{15}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-\frac{15}{2}\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-\frac{15}{2}\right)}}{2}$$

Multiply -4 times -\frac{15}{2}.

$$x=\frac{-1±\sqrt{1+30}}{2}$$

Add 1 to 30.

$$x=\frac{-1±\sqrt{31}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{31}}{2} when ± is plus. Add -1 to \sqrt{31}.

$$x=\frac{\sqrt{31}-1}{2}$$

Now solve the equation x=\frac{-1±\sqrt{31}}{2} when ± is minus. Subtract \sqrt{31} from -1.

$$x=\frac{-\sqrt{31}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{31}-1}{2} x=\frac{-\sqrt{31}-1}{2}$$

Steps for Completing the Square

Divide 9 by \frac{6}{5} by multiplying 9 by the reciprocal of \frac{6}{5}.

$$9\times \frac{5}{6}=x^{2}-2x+3x$$

Multiply 9 and \frac{5}{6} to get \frac{15}{2}.

$$\frac{15}{2}=x^{2}-2x+3x$$

Combine -2x and 3x to get x.

$$\frac{15}{2}=x^{2}+x$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+x=\frac{15}{2}$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{15}{2}+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{15}{2}+\frac{1}{4}$$

Add \frac{15}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{31}{4}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{31}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{31}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{31}}{2} x+\frac{1}{2}=-\frac{\sqrt{31}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{31}-1}{2} x=\frac{-\sqrt{31}-1}{2}$$