Solve 9x^2+150x-119=0

$$9 { x }^{ 2 } +150x-119=0$$

Solve for x

$$x=\frac{2\sqrt{186}-25}{3}\approx 0.758787798$$

$$x=\frac{-2\sqrt{186}-25}{3}\approx -17.425454465$$

Steps Using the Quadratic Formula

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.

$$9x^{2}+150x-119=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 150 for b, and -119 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-150±\sqrt{150^{2}-4\times 9\left(-119\right)}}{2\times 9}$$

Square 150.

$$x=\frac{-150±\sqrt{22500-4\times 9\left(-119\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-150±\sqrt{22500-36\left(-119\right)}}{2\times 9}$$

Multiply -36 times -119.

$$x=\frac{-150±\sqrt{22500+4284}}{2\times 9}$$

Add 22500 to 4284.

$$x=\frac{-150±\sqrt{26784}}{2\times 9}$$

Take the square root of 26784.

$$x=\frac{-150±12\sqrt{186}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-150±12\sqrt{186}}{18}$$

Now solve the equation x=\frac{-150±12\sqrt{186}}{18} when ± is plus. Add -150 to 12\sqrt{186}.

$$x=\frac{12\sqrt{186}-150}{18}$$

Divide -150+12\sqrt{186} by 18.

$$x=\frac{2\sqrt{186}-25}{3}$$

Now solve the equation x=\frac{-150±12\sqrt{186}}{18} when ± is minus. Subtract 12\sqrt{186} from -150.

$$x=\frac{-12\sqrt{186}-150}{18}$$

Divide -150-12\sqrt{186} by 18.

$$x=\frac{-2\sqrt{186}-25}{3}$$

The equation is now solved.

$$x=\frac{2\sqrt{186}-25}{3} x=\frac{-2\sqrt{186}-25}{3}$$

Steps for Completing the Square

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.

$$9x^{2}+150x-119=0$$

Add 119 to both sides of the equation.

$$9x^{2}+150x-119-\left(-119\right)=-\left(-119\right)$$

Subtracting -119 from itself leaves 0.

$$9x^{2}+150x=-\left(-119\right)$$

Subtract -119 from 0.

$$9x^{2}+150x=119$$

Divide both sides by 9.

$$\frac{9x^{2}+150x}{9}=\frac{119}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{150}{9}x=\frac{119}{9}$$

Reduce the fraction \frac{150}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{50}{3}x=\frac{119}{9}$$

Divide \frac{50}{3}, the coefficient of the x term, by 2 to get \frac{25}{3}. Then add the square of \frac{25}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{50}{3}x+\left(\frac{25}{3}\right)^{2}=\frac{119}{9}+\left(\frac{25}{3}\right)^{2}$$

Square \frac{25}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{50}{3}x+\frac{625}{9}=\frac{119+625}{9}$$

Add \frac{119}{9} to \frac{625}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{50}{3}x+\frac{625}{9}=\frac{248}{3}$$

Factor x^{2}+\frac{50}{3}x+\frac{625}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{25}{3}\right)^{2}=\frac{248}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{25}{3}\right)^{2}}=\sqrt{\frac{248}{3}}$$

Simplify.

$$x+\frac{25}{3}=\frac{2\sqrt{186}}{3} x+\frac{25}{3}=-\frac{2\sqrt{186}}{3}$$

Subtract \frac{25}{3} from both sides of the equation.

$$x=\frac{2\sqrt{186}-25}{3} x=\frac{-2\sqrt{186}-25}{3}$$