Solve -3x^2-33x+126=0

$$-3 { x }^{ 2 } -33x+126=0$$

Solve for x

$$x=-14$$

$$x=3$$

Steps Using Factoring By Grouping

Divide both sides by 3.

$$-x^{2}-11x+42=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+42. To find a and b, set up a system to be solved.

$$a+b=-11 ab=-42=-42$$

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 -42.

$$1,-42 2,-21 3,-14 6,-7$$

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -11.

$$a=3 b=-14$$

Rewrite -x^{2}-11x+42 as \left(-x^{2}+3x\right)+\left(-14x+42\right).

$$\left(-x^{2}+3x\right)+\left(-14x+42\right)$$

Factor out x in the first and 14 in the second group.

$$x\left(-x+3\right)+14\left(-x+3\right)$$

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

$$\left(-x+3\right)\left(x+14\right)$$

To find equation solutions, solve -x+3=0 and x+14=0.

$$x=3 x=-14$$

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.

$$-3x^{2}-33x+126=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -33 for b, and 126 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\left(-3\right)\times 126}}{2\left(-3\right)}$$

Square -33.

$$x=\frac{-\left(-33\right)±\sqrt{1089-4\left(-3\right)\times 126}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-33\right)±\sqrt{1089+12\times 126}}{2\left(-3\right)}$$

Multiply 12 times 126.

$$x=\frac{-\left(-33\right)±\sqrt{1089+1512}}{2\left(-3\right)}$$

Add 1089 to 1512.

$$x=\frac{-\left(-33\right)±\sqrt{2601}}{2\left(-3\right)}$$

Take the square root of 2601.

$$x=\frac{-\left(-33\right)±51}{2\left(-3\right)}$$

The opposite of -33 is 33.

$$x=\frac{33±51}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{33±51}{-6}$$

Now solve the equation x=\frac{33±51}{-6} when ± is plus. Add 33 to 51.

$$x=\frac{84}{-6}$$

Divide 84 by -6.

$$x=-14$$

Now solve the equation x=\frac{33±51}{-6} when ± is minus. Subtract 51 from 33.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

The equation is now solved.

$$x=-14 x=3$$

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.

$$-3x^{2}-33x+126=0$$

Subtract 126 from both sides of the equation.

$$-3x^{2}-33x+126-126=-126$$

Subtracting 126 from itself leaves 0.

$$-3x^{2}-33x=-126$$

Divide both sides by -3.

$$\frac{-3x^{2}-33x}{-3}=-\frac{126}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{33}{-3}\right)x=-\frac{126}{-3}$$

Divide -33 by -3.

$$x^{2}+11x=-\frac{126}{-3}$$

Divide -126 by -3.

$$x^{2}+11x=42$$

Divide 11, the coefficient of the x term, by 2 to get \frac{11}{2}. Then add the square of \frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+11x+\left(\frac{11}{2}\right)^{2}=42+\left(\frac{11}{2}\right)^{2}$$

Square \frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+11x+\frac{121}{4}=42+\frac{121}{4}$$

Add 42 to \frac{121}{4}.

$$x^{2}+11x+\frac{121}{4}=\frac{289}{4}$$

Factor x^{2}+11x+\frac{121}{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}.

$$\left(x+\frac{11}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$x+\frac{11}{2}=\frac{17}{2} x+\frac{11}{2}=-\frac{17}{2}$$

Subtract \frac{11}{2} from both sides of the equation.

$$x=3 x=-14$$