Solve -48=-4x^2+4x

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

Solve for x

$$x=-3$$

$$x=4$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

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

Add 48 to both sides.

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

Divide both sides by 4.

$$-x^{2}+x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=1 ab=-12=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

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

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=4 b=-3$$

Rewrite -x^{2}+x+12 as \left(-x^{2}+4x\right)+\left(-3x+12\right).

$$\left(-x^{2}+4x\right)+\left(-3x+12\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(x-4\right)-3\left(x-4\right)$$

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

$$\left(x-4\right)\left(-x-3\right)$$

To find equation solutions, solve x-4=0 and -x-3=0.

$$x=4 x=-3$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

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

Add 48 to both sides.

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

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

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

Square 4.

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

Multiply -4 times -4.

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

Multiply 16 times 48.

$$x=\frac{-4±\sqrt{16+768}}{2\left(-4\right)}$$

Add 16 to 768.

$$x=\frac{-4±\sqrt{784}}{2\left(-4\right)}$$

Take the square root of 784.

$$x=\frac{-4±28}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-4±28}{-8}$$

Now solve the equation x=\frac{-4±28}{-8} when ± is plus. Add -4 to 28.

$$x=\frac{24}{-8}$$

Divide 24 by -8.

$$x=-3$$

Now solve the equation x=\frac{-4±28}{-8} when ± is minus. Subtract 28 from -4.

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

Divide -32 by -8.

$$x=4$$

The equation is now solved.

$$x=-3 x=4$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

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

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

Divide 4 by -4.

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

Divide -48 by -4.

$$x^{2}-x=12$$

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=12+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{49}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{7}{2} x-\frac{1}{2}=-\frac{7}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=4 x=-3$$