Solve -4x^2/3+8x/3+4=0

$$\frac{ -4 { x }^{ 2 } }{ 3 } + \frac{ 8x }{ 3 } +4=0$$

Solve for x

$$x=-1$$

$$x=3$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 3.

$$-4x^{2}+8x+12=0$$

Divide both sides by 4.

$$-x^{2}+2x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3=-3$$

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. The only such pair is the system solution.

$$a=3 b=-1$$

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

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

Factor out -x in the first and -1 in the second group.

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

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x-1\right)$$

To find equation solutions, solve x-3=0 and -x-1=0.

$$x=3 x=-1$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

$$-4x^{2}+8x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 8 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-4\right)\times 12}}{2\left(-4\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-4\right)\times 12}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-8±\sqrt{64+16\times 12}}{2\left(-4\right)}$$

Multiply 16 times 12.

$$x=\frac{-8±\sqrt{64+192}}{2\left(-4\right)}$$

Add 64 to 192.

$$x=\frac{-8±\sqrt{256}}{2\left(-4\right)}$$

Take the square root of 256.

$$x=\frac{-8±16}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-8±16}{-8}$$

Now solve the equation x=\frac{-8±16}{-8} when ± is plus. Add -8 to 16.

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

Divide 8 by -8.

$$x=-1$$

Now solve the equation x=\frac{-8±16}{-8} when ± is minus. Subtract 16 from -8.

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

Divide -24 by -8.

$$x=3$$

The equation is now solved.

$$x=-1 x=3$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

$$-4x^{2}+8x+12=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$-4x^{2}+8x=-12$$

Divide both sides by -4.

$$\frac{-4x^{2}+8x}{-4}=-\frac{12}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{8}{-4}x=-\frac{12}{-4}$$

Divide 8 by -4.

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

Divide -12 by -4.

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

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.

$$x^{2}-2x+1=3+1$$

Add 3 to 1.

$$x^{2}-2x+1=4$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=2 x-1=-2$$

Add 1 to both sides of the equation.

$$x=3 x=-1$$