Solve 4x^2-32x=148

$$4 x ^ { 2 } - 32 x = 148$$

Solve for x

$$x=\sqrt{53}+4\approx 11.280109889$$

$$x=4-\sqrt{53}\approx -3.280109889$$

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.

$$4x^{2}-32x=148$$

Subtract 148 from both sides of the equation.

$$4x^{2}-32x-148=148-148$$

Subtracting 148 from itself leaves 0.

$$4x^{2}-32x-148=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -32 for b, and -148 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 4\left(-148\right)}}{2\times 4}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 4\left(-148\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-32\right)±\sqrt{1024-16\left(-148\right)}}{2\times 4}$$

Multiply -16 times -148.

$$x=\frac{-\left(-32\right)±\sqrt{1024+2368}}{2\times 4}$$

Add 1024 to 2368.

$$x=\frac{-\left(-32\right)±\sqrt{3392}}{2\times 4}$$

Take the square root of 3392.

$$x=\frac{-\left(-32\right)±8\sqrt{53}}{2\times 4}$$

The opposite of -32 is 32.

$$x=\frac{32±8\sqrt{53}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{32±8\sqrt{53}}{8}$$

Now solve the equation x=\frac{32±8\sqrt{53}}{8} when ± is plus. Add 32 to 8\sqrt{53}.

$$x=\frac{8\sqrt{53}+32}{8}$$

Divide 32+8\sqrt{53} by 8.

$$x=\sqrt{53}+4$$

Now solve the equation x=\frac{32±8\sqrt{53}}{8} when ± is minus. Subtract 8\sqrt{53} from 32.

$$x=\frac{32-8\sqrt{53}}{8}$$

Divide 32-8\sqrt{53} by 8.

$$x=4-\sqrt{53}$$

The equation is now solved.

$$x=\sqrt{53}+4 x=4-\sqrt{53}$$

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.

$$4x^{2}-32x=148$$

Divide both sides by 4.

$$\frac{4x^{2}-32x}{4}=\frac{148}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{32}{4}\right)x=\frac{148}{4}$$

Divide -32 by 4.

$$x^{2}-8x=\frac{148}{4}$$

Divide 148 by 4.

$$x^{2}-8x=37$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=37+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=37+16$$

Add 37 to 16.

$$x^{2}-8x+16=53$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=53$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{53}$$

Simplify.

$$x-4=\sqrt{53} x-4=-\sqrt{53}$$

Add 4 to both sides of the equation.

$$x=\sqrt{53}+4 x=4-\sqrt{53}$$