Solve 9x^2-35=-6x

$$9 x ^ { 2 } - 35 = - 6 x$$

Solve for x

$$x = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using Factoring By Grouping

Add 6x to both sides.

$$9x^{2}-35+6x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$9x^{2}+6x-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=6 ab=9\left(-35\right)=-315$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -315.

$$-1,315 -3,105 -5,63 -7,45 -9,35 -15,21$$

Calculate the sum for each pair.

$$-1+315=314 -3+105=102 -5+63=58 -7+45=38 -9+35=26 -15+21=6$$

The solution is the pair that gives sum 6.

$$a=-15 b=21$$

Rewrite 9x^{2}+6x-35 as \left(9x^{2}-15x\right)+\left(21x-35\right).

$$\left(9x^{2}-15x\right)+\left(21x-35\right)$$

Factor out 3x in the first and 7 in the second group.

$$3x\left(3x-5\right)+7\left(3x-5\right)$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(3x+7\right)$$

To find equation solutions, solve 3x-5=0 and 3x+7=0.

$$x=\frac{5}{3} x=-\frac{7}{3}$$

Steps Using the Quadratic Formula

Add 6x to both sides.

$$9x^{2}-35+6x=0$$

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.

$$9x^{2}+6x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 6 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\left(-35\right)}}{2\times 9}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 9\left(-35\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-6±\sqrt{36-36\left(-35\right)}}{2\times 9}$$

Multiply -36 times -35.

$$x=\frac{-6±\sqrt{36+1260}}{2\times 9}$$

Add 36 to 1260.

$$x=\frac{-6±\sqrt{1296}}{2\times 9}$$

Take the square root of 1296.

$$x=\frac{-6±36}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-6±36}{18}$$

Now solve the equation x=\frac{-6±36}{18} when ± is plus. Add -6 to 36.

$$x=\frac{30}{18}$$

Reduce the fraction \frac{30}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{-6±36}{18} when ± is minus. Subtract 36 from -6.

$$x=-\frac{42}{18}$$

Reduce the fraction \frac{-42}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{7}{3}$$

The equation is now solved.

$$x=\frac{5}{3} x=-\frac{7}{3}$$

Steps for Completing the Square

Add 6x to both sides.

$$9x^{2}-35+6x=0$$

Add 35 to both sides. Anything plus zero gives itself.

$$9x^{2}+6x=35$$

Divide both sides by 9.

$$\frac{9x^{2}+6x}{9}=\frac{35}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{6}{9}x=\frac{35}{9}$$

Reduce the fraction \frac{6}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{2}{3}x=\frac{35}{9}$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{35}{9}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{35+1}{9}$$

Add \frac{35}{9} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=4$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{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}.

$$\left(x+\frac{1}{3}\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+\frac{1}{3}=2 x+\frac{1}{3}=-2$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=\frac{5}{3} x=-\frac{7}{3}$$