Solve 3x^2-18x+5=47

$$3 x ^ { 2 } - 18 x + 5 = 47$$

Solve for x

$$x=\sqrt{23}+3\approx 7.795831523$$

$$x=3-\sqrt{23}\approx -1.795831523$$

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.

$$3x^{2}-18x+5=47$$

Subtract 47 from both sides of the equation.

$$3x^{2}-18x+5-47=47-47$$

Subtracting 47 from itself leaves 0.

$$3x^{2}-18x+5-47=0$$

Subtract 47 from 5.

$$3x^{2}-18x-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -18 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\left(-42\right)}}{2\times 3}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 3\left(-42\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-18\right)±\sqrt{324-12\left(-42\right)}}{2\times 3}$$

Multiply -12 times -42.

$$x=\frac{-\left(-18\right)±\sqrt{324+504}}{2\times 3}$$

Add 324 to 504.

$$x=\frac{-\left(-18\right)±\sqrt{828}}{2\times 3}$$

Take the square root of 828.

$$x=\frac{-\left(-18\right)±6\sqrt{23}}{2\times 3}$$

The opposite of -18 is 18.

$$x=\frac{18±6\sqrt{23}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{18±6\sqrt{23}}{6}$$

Now solve the equation x=\frac{18±6\sqrt{23}}{6} when ± is plus. Add 18 to 6\sqrt{23}.

$$x=\frac{6\sqrt{23}+18}{6}$$

Divide 18+6\sqrt{23} by 6.

$$x=\sqrt{23}+3$$

Now solve the equation x=\frac{18±6\sqrt{23}}{6} when ± is minus. Subtract 6\sqrt{23} from 18.

$$x=\frac{18-6\sqrt{23}}{6}$$

Divide 18-6\sqrt{23} by 6.

$$x=3-\sqrt{23}$$

The equation is now solved.

$$x=\sqrt{23}+3 x=3-\sqrt{23}$$

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.

$$3x^{2}-18x+5=47$$

Subtract 5 from both sides of the equation.

$$3x^{2}-18x+5-5=47-5$$

Subtracting 5 from itself leaves 0.

$$3x^{2}-18x=47-5$$

Subtract 5 from 47.

$$3x^{2}-18x=42$$

Divide both sides by 3.

$$\frac{3x^{2}-18x}{3}=\frac{42}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{18}{3}\right)x=\frac{42}{3}$$

Divide -18 by 3.

$$x^{2}-6x=\frac{42}{3}$$

Divide 42 by 3.

$$x^{2}-6x=14$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=14+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=14+9$$

Add 14 to 9.

$$x^{2}-6x+9=23$$

Factor x^{2}-6x+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}.

$$\left(x-3\right)^{2}=23$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{23}$$

Simplify.

$$x-3=\sqrt{23} x-3=-\sqrt{23}$$

Add 3 to both sides of the equation.

$$x=\sqrt{23}+3 x=3-\sqrt{23}$$