Solve 9x^2=3-7x

$$9 x ^ { 2 } = 3 - 7 x$$

Solve for x

$$x=\frac{\sqrt{157}-7}{18}\approx 0.307220227$$

$$x=\frac{-\sqrt{157}-7}{18}\approx -1.084998005$$

Steps Using the Quadratic Formula

Subtract 3 from both sides.

$$9x^{2}-3=-7x$$

Add 7x to both sides.

$$9x^{2}-3+7x=0$$

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.

$$9x^{2}+7x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 7 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 9\left(-3\right)}}{2\times 9}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 9\left(-3\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-7±\sqrt{49-36\left(-3\right)}}{2\times 9}$$

Multiply -36 times -3.

$$x=\frac{-7±\sqrt{49+108}}{2\times 9}$$

Add 49 to 108.

$$x=\frac{-7±\sqrt{157}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-7±\sqrt{157}}{18}$$

Now solve the equation x=\frac{-7±\sqrt{157}}{18} when ± is plus. Add -7 to \sqrt{157}.

$$x=\frac{\sqrt{157}-7}{18}$$

Now solve the equation x=\frac{-7±\sqrt{157}}{18} when ± is minus. Subtract \sqrt{157} from -7.

$$x=\frac{-\sqrt{157}-7}{18}$$

The equation is now solved.

$$x=\frac{\sqrt{157}-7}{18} x=\frac{-\sqrt{157}-7}{18}$$

Steps for Completing the Square

Add 7x to both sides.

$$9x^{2}+7x=3$$

Divide both sides by 9.

$$\frac{9x^{2}+7x}{9}=\frac{3}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{7}{9}x=\frac{3}{9}$$

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

$$x^{2}+\frac{7}{9}x=\frac{1}{3}$$

Divide \frac{7}{9}, the coefficient of the x term, by 2 to get \frac{7}{18}. Then add the square of \frac{7}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{9}x+\left(\frac{7}{18}\right)^{2}=\frac{1}{3}+\left(\frac{7}{18}\right)^{2}$$

Square \frac{7}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{9}x+\frac{49}{324}=\frac{1}{3}+\frac{49}{324}$$

Add \frac{1}{3} to \frac{49}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{9}x+\frac{49}{324}=\frac{157}{324}$$

Factor x^{2}+\frac{7}{9}x+\frac{49}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{18}\right)^{2}=\frac{157}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{18}\right)^{2}}=\sqrt{\frac{157}{324}}$$

Simplify.

$$x+\frac{7}{18}=\frac{\sqrt{157}}{18} x+\frac{7}{18}=-\frac{\sqrt{157}}{18}$$

Subtract \frac{7}{18} from both sides of the equation.

$$x=\frac{\sqrt{157}-7}{18} x=\frac{-\sqrt{157}-7}{18}$$