Solve 8x^2-36x+4x-48=0

$$8 x ^ { 2 } - 36 x + 4 x - 48 = 0$$

Solve for x

$$x=\sqrt{10}+2\approx 5.16227766$$

$$x=2-\sqrt{10}\approx -1.16227766$$

Steps Using the Quadratic Formula

Combine -36x and 4x to get -32x.

$$8x^{2}-32x-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -32 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 8\left(-48\right)}}{2\times 8}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 8\left(-48\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-32\right)±\sqrt{1024-32\left(-48\right)}}{2\times 8}$$

Multiply -32 times -48.

$$x=\frac{-\left(-32\right)±\sqrt{1024+1536}}{2\times 8}$$

Add 1024 to 1536.

$$x=\frac{-\left(-32\right)±\sqrt{2560}}{2\times 8}$$

Take the square root of 2560.

$$x=\frac{-\left(-32\right)±16\sqrt{10}}{2\times 8}$$

The opposite of -32 is 32.

$$x=\frac{32±16\sqrt{10}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{32±16\sqrt{10}}{16}$$

Now solve the equation x=\frac{32±16\sqrt{10}}{16} when ± is plus. Add 32 to 16\sqrt{10}.

$$x=\frac{16\sqrt{10}+32}{16}$$

Divide 32+16\sqrt{10} by 16.

$$x=\sqrt{10}+2$$

Now solve the equation x=\frac{32±16\sqrt{10}}{16} when ± is minus. Subtract 16\sqrt{10} from 32.

$$x=\frac{32-16\sqrt{10}}{16}$$

Divide 32-16\sqrt{10} by 16.

$$x=2-\sqrt{10}$$

The equation is now solved.

$$x=\sqrt{10}+2 x=2-\sqrt{10}$$

Steps for Completing the Square

Combine -36x and 4x to get -32x.

$$8x^{2}-32x-48=0$$

Add 48 to both sides. Anything plus zero gives itself.

$$8x^{2}-32x=48$$

Divide both sides by 8.

$$\frac{8x^{2}-32x}{8}=\frac{48}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{32}{8}\right)x=\frac{48}{8}$$

Divide -32 by 8.

$$x^{2}-4x=\frac{48}{8}$$

Divide 48 by 8.

$$x^{2}-4x=6$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=6+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=6+4$$

Add 6 to 4.

$$x^{2}-4x+4=10$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=10$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{10}$$

Simplify.

$$x-2=\sqrt{10} x-2=-\sqrt{10}$$

Add 2 to both sides of the equation.

$$x=\sqrt{10}+2 x=2-\sqrt{10}$$