Solve -2x^2+48=4x

$$- 2 x ^ { 2 } + 48 = 4 x$$

Solve for x

$$x=-6$$

$$x=4$$

Steps Using Factoring By Grouping

Subtract 4x from both sides.

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

Divide both sides by 2.

$$-x^{2}+24-2x=0$$

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

$$-x^{2}-2x+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-24=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -2.

$$a=4 b=-6$$

Rewrite -x^{2}-2x+24 as \left(-x^{2}+4x\right)+\left(-6x+24\right).

$$\left(-x^{2}+4x\right)+\left(-6x+24\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(-x+4\right)+6\left(-x+4\right)$$

Factor out common term -x+4 by using distributive property.

$$\left(-x+4\right)\left(x+6\right)$$

To find equation solutions, solve -x+4=0 and x+6=0.

$$x=4 x=-6$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

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

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.

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

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-2\right)\times 48}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-4\right)±\sqrt{16+8\times 48}}{2\left(-2\right)}$$

Multiply 8 times 48.

$$x=\frac{-\left(-4\right)±\sqrt{16+384}}{2\left(-2\right)}$$

Add 16 to 384.

$$x=\frac{-\left(-4\right)±\sqrt{400}}{2\left(-2\right)}$$

Take the square root of 400.

$$x=\frac{-\left(-4\right)±20}{2\left(-2\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±20}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{4±20}{-4}$$

Now solve the equation x=\frac{4±20}{-4} when ± is plus. Add 4 to 20.

$$x=\frac{24}{-4}$$

Divide 24 by -4.

$$x=-6$$

Now solve the equation x=\frac{4±20}{-4} when ± is minus. Subtract 20 from 4.

$$x=-\frac{16}{-4}$$

Divide -16 by -4.

$$x=4$$

The equation is now solved.

$$x=-6 x=4$$

Steps for Completing the Square

Subtract 4x from both sides.

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

Subtract 48 from both sides. Anything subtracted from zero gives its negation.

$$-2x^{2}-4x=-48$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{4}{-2}\right)x=-\frac{48}{-2}$$

Divide -4 by -2.

$$x^{2}+2x=-\frac{48}{-2}$$

Divide -48 by -2.

$$x^{2}+2x=24$$

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

$$x^{2}+2x+1^{2}=24+1^{2}$$

Square 1.

$$x^{2}+2x+1=24+1$$

Add 24 to 1.

$$x^{2}+2x+1=25$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{25}$$

Simplify.

$$x+1=5 x+1=-5$$

Subtract 1 from both sides of the equation.

$$x=4 x=-6$$