Solve 18x^2+66x-126/17-3x-6-18=0

$$\frac { 18 x ^ { 2 } + 66 x - 126 } { 17 - 3 x } - 6 - 18 = 0$$

Solve for x

$$x = \frac{\sqrt{1597} - 23}{6} \approx 2.827080401$$

$$x=\frac{-\sqrt{1597}-23}{6}\approx -10.493747068$$

Steps Using the Quadratic Formula

Variable x cannot be equal to \frac{17}{3} since division by zero is not defined. Multiply both sides of the equation by -3x+17.

$$18x^{2}+66x-126+\left(-3x+17\right)\left(-6\right)+\left(-3x+17\right)\left(-18\right)=0$$

Use the distributive property to multiply -3x+17 by -6.

$$18x^{2}+66x-126+18x-102+\left(-3x+17\right)\left(-18\right)=0$$

Combine 66x and 18x to get 84x.

$$18x^{2}+84x-126-102+\left(-3x+17\right)\left(-18\right)=0$$

Subtract 102 from -126 to get -228.

$$18x^{2}+84x-228+\left(-3x+17\right)\left(-18\right)=0$$

Use the distributive property to multiply -3x+17 by -18.

$$18x^{2}+84x-228+54x-306=0$$

Combine 84x and 54x to get 138x.

$$18x^{2}+138x-228-306=0$$

Subtract 306 from -228 to get -534.

$$18x^{2}+138x-534=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, 138 for b, and -534 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-138±\sqrt{138^{2}-4\times 18\left(-534\right)}}{2\times 18}$$

Square 138.

$$x=\frac{-138±\sqrt{19044-4\times 18\left(-534\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-138±\sqrt{19044-72\left(-534\right)}}{2\times 18}$$

Multiply -72 times -534.

$$x=\frac{-138±\sqrt{19044+38448}}{2\times 18}$$

Add 19044 to 38448.

$$x=\frac{-138±\sqrt{57492}}{2\times 18}$$

Take the square root of 57492.

$$x=\frac{-138±6\sqrt{1597}}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-138±6\sqrt{1597}}{36}$$

Now solve the equation x=\frac{-138±6\sqrt{1597}}{36} when ± is plus. Add -138 to 6\sqrt{1597}.

$$x=\frac{6\sqrt{1597}-138}{36}$$

Divide -138+6\sqrt{1597} by 36.

$$x=\frac{\sqrt{1597}-23}{6}$$

Now solve the equation x=\frac{-138±6\sqrt{1597}}{36} when ± is minus. Subtract 6\sqrt{1597} from -138.

$$x=\frac{-6\sqrt{1597}-138}{36}$$

Divide -138-6\sqrt{1597} by 36.

$$x=\frac{-\sqrt{1597}-23}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{1597}-23}{6} x=\frac{-\sqrt{1597}-23}{6}$$

Steps for Completing the Square

Variable x cannot be equal to \frac{17}{3} since division by zero is not defined. Multiply both sides of the equation by -3x+17.

$$18x^{2}+66x-126+\left(-3x+17\right)\left(-6\right)+\left(-3x+17\right)\left(-18\right)=0$$

Use the distributive property to multiply -3x+17 by -6.

$$18x^{2}+66x-126+18x-102+\left(-3x+17\right)\left(-18\right)=0$$

Combine 66x and 18x to get 84x.

$$18x^{2}+84x-126-102+\left(-3x+17\right)\left(-18\right)=0$$

Subtract 102 from -126 to get -228.

$$18x^{2}+84x-228+\left(-3x+17\right)\left(-18\right)=0$$

Use the distributive property to multiply -3x+17 by -18.

$$18x^{2}+84x-228+54x-306=0$$

Combine 84x and 54x to get 138x.

$$18x^{2}+138x-228-306=0$$

Subtract 306 from -228 to get -534.

$$18x^{2}+138x-534=0$$

Add 534 to both sides. Anything plus zero gives itself.

$$18x^{2}+138x=534$$

Divide both sides by 18.

$$\frac{18x^{2}+138x}{18}=\frac{534}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\frac{138}{18}x=\frac{534}{18}$$

Reduce the fraction \frac{138}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{23}{3}x=\frac{534}{18}$$

Reduce the fraction \frac{534}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{23}{3}x=\frac{89}{3}$$

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.

$$x^{2}+\frac{23}{3}x+\left(\frac{23}{6}\right)^{2}=\frac{89}{3}+\left(\frac{23}{6}\right)^{2}$$

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

$$x^{2}+\frac{23}{3}x+\frac{529}{36}=\frac{89}{3}+\frac{529}{36}$$

Add \frac{89}{3} to \frac{529}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{23}{3}x+\frac{529}{36}=\frac{1597}{36}$$

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}.

$$\left(x+\frac{23}{6}\right)^{2}=\frac{1597}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{23}{6}\right)^{2}}=\sqrt{\frac{1597}{36}}$$

Simplify.

$$x+\frac{23}{6}=\frac{\sqrt{1597}}{6} x+\frac{23}{6}=-\frac{\sqrt{1597}}{6}$$

Subtract \frac{23}{6} from both sides of the equation.

$$x=\frac{\sqrt{1597}-23}{6} x=\frac{-\sqrt{1597}-23}{6}$$