Solve 3-3x^2=16-32x+16x^2

$$3 - 3 x ^ { 2 } = 16 - 32 x + 16 x ^ { 2 }$$

Solve for x

$$x=1$$

$$x=\frac{13}{19}\approx 0.684210526$$

Steps Using Factoring By Grouping

Subtract 16 from both sides.

$$3-3x^{2}-16=-32x+16x^{2}$$

Subtract 16 from 3 to get -13.

$$-13-3x^{2}=-32x+16x^{2}$$

Add 32x to both sides.

$$-13-3x^{2}+32x=16x^{2}$$

Subtract 16x^{2} from both sides.

$$-13-3x^{2}+32x-16x^{2}=0$$

Combine -3x^{2} and -16x^{2} to get -19x^{2}.

$$-13-19x^{2}+32x=0$$

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

$$-19x^{2}+32x-13=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -19x^{2}+ax+bx-13. To find a and b, set up a system to be solved.

$$a+b=32 ab=-19\left(-13\right)=247$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 247.

$$1,247 13,19$$

Calculate the sum for each pair.

$$1+247=248 13+19=32$$

The solution is the pair that gives sum 32.

$$a=19 b=13$$

Rewrite -19x^{2}+32x-13 as \left(-19x^{2}+19x\right)+\left(13x-13\right).

$$\left(-19x^{2}+19x\right)+\left(13x-13\right)$$

Factor out 19x in the first and -13 in the second group.

$$19x\left(-x+1\right)-13\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(19x-13\right)$$

To find equation solutions, solve -x+1=0 and 19x-13=0.

$$x=1 x=\frac{13}{19}$$

Steps Using the Quadratic Formula

Subtract 16 from both sides.

$$3-3x^{2}-16=-32x+16x^{2}$$

Subtract 16 from 3 to get -13.

$$-13-3x^{2}=-32x+16x^{2}$$

Add 32x to both sides.

$$-13-3x^{2}+32x=16x^{2}$$

Subtract 16x^{2} from both sides.

$$-13-3x^{2}+32x-16x^{2}=0$$

Combine -3x^{2} and -16x^{2} to get -19x^{2}.

$$-13-19x^{2}+32x=0$$

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.

$$-19x^{2}+32x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -19 for a, 32 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}-4\left(-19\right)\left(-13\right)}}{2\left(-19\right)}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\left(-19\right)\left(-13\right)}}{2\left(-19\right)}$$

Multiply -4 times -19.

$$x=\frac{-32±\sqrt{1024+76\left(-13\right)}}{2\left(-19\right)}$$

Multiply 76 times -13.

$$x=\frac{-32±\sqrt{1024-988}}{2\left(-19\right)}$$

Add 1024 to -988.

$$x=\frac{-32±\sqrt{36}}{2\left(-19\right)}$$

Take the square root of 36.

$$x=\frac{-32±6}{2\left(-19\right)}$$

Multiply 2 times -19.

$$x=\frac{-32±6}{-38}$$

Now solve the equation x=\frac{-32±6}{-38} when ± is plus. Add -32 to 6.

$$x=-\frac{26}{-38}$$

Reduce the fraction \frac{-26}{-38} to lowest terms by extracting and canceling out 2.

$$x=\frac{13}{19}$$

Now solve the equation x=\frac{-32±6}{-38} when ± is minus. Subtract 6 from -32.

$$x=-\frac{38}{-38}$$

Divide -38 by -38.

$$x=1$$

The equation is now solved.

$$x=\frac{13}{19} x=1$$

Steps for Completing the Square

Add 32x to both sides.

$$3-3x^{2}+32x=16+16x^{2}$$

Subtract 16x^{2} from both sides.

$$3-3x^{2}+32x-16x^{2}=16$$

Combine -3x^{2} and -16x^{2} to get -19x^{2}.

$$3-19x^{2}+32x=16$$

Subtract 3 from both sides.

$$-19x^{2}+32x=16-3$$

Subtract 3 from 16 to get 13.

$$-19x^{2}+32x=13$$

Divide both sides by -19.

$$\frac{-19x^{2}+32x}{-19}=\frac{13}{-19}$$

Dividing by -19 undoes the multiplication by -19.

$$x^{2}+\frac{32}{-19}x=\frac{13}{-19}$$

Divide 32 by -19.

$$x^{2}-\frac{32}{19}x=\frac{13}{-19}$$

Divide 13 by -19.

$$x^{2}-\frac{32}{19}x=-\frac{13}{19}$$

Divide -\frac{32}{19}, the coefficient of the x term, by 2 to get -\frac{16}{19}. Then add the square of -\frac{16}{19} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{19}x+\left(-\frac{16}{19}\right)^{2}=-\frac{13}{19}+\left(-\frac{16}{19}\right)^{2}$$

Square -\frac{16}{19} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{19}x+\frac{256}{361}=-\frac{13}{19}+\frac{256}{361}$$

Add -\frac{13}{19} to \frac{256}{361} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{32}{19}x+\frac{256}{361}=\frac{9}{361}$$

Factor x^{2}-\frac{32}{19}x+\frac{256}{361}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{16}{19}\right)^{2}=\frac{9}{361}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{19}\right)^{2}}=\sqrt{\frac{9}{361}}$$

Simplify.

$$x-\frac{16}{19}=\frac{3}{19} x-\frac{16}{19}=-\frac{3}{19}$$

Add \frac{16}{19} to both sides of the equation.

$$x=1 x=\frac{13}{19}$$