Solve -3x^2-44x-97=29-5x

$$-3 { x }^{ 2 } -44x-97 = 29-5x$$

Solve for x

$$x=-7$$

$$x=-6$$

Steps Using Factoring By Grouping

Subtract 29 from both sides.

$$-3x^{2}-44x-97-29=-5x$$

Subtract 29 from -97 to get -126.

$$-3x^{2}-44x-126=-5x$$

Add 5x to both sides.

$$-3x^{2}-44x-126+5x=0$$

Combine -44x and 5x to get -39x.

$$-3x^{2}-39x-126=0$$

Divide both sides by 3.

$$-x^{2}-13x-42=0$$

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.

$$a+b=-13 ab=-\left(-42\right)=42$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 42.

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

Calculate the sum for each pair.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

The solution is the pair that gives sum -13.

$$a=-6 b=-7$$

Rewrite -x^{2}-13x-42 as \left(-x^{2}-6x\right)+\left(-7x-42\right).

$$\left(-x^{2}-6x\right)+\left(-7x-42\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(-x-6\right)+7\left(-x-6\right)$$

Factor out common term -x-6 by using distributive property.

$$\left(-x-6\right)\left(x+7\right)$$

To find equation solutions, solve -x-6=0 and x+7=0.

$$x=-6 x=-7$$

Steps Using the Quadratic Formula

Subtract 29 from both sides.

$$-3x^{2}-44x-97-29=-5x$$

Subtract 29 from -97 to get -126.

$$-3x^{2}-44x-126=-5x$$

Add 5x to both sides.

$$-3x^{2}-44x-126+5x=0$$

Combine -44x and 5x to get -39x.

$$-3x^{2}-39x-126=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -39 for b, and -126 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}-4\left(-3\right)\left(-126\right)}}{2\left(-3\right)}$$

Square -39.

$$x=\frac{-\left(-39\right)±\sqrt{1521-4\left(-3\right)\left(-126\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-39\right)±\sqrt{1521+12\left(-126\right)}}{2\left(-3\right)}$$

Multiply 12 times -126.

$$x=\frac{-\left(-39\right)±\sqrt{1521-1512}}{2\left(-3\right)}$$

Add 1521 to -1512.

$$x=\frac{-\left(-39\right)±\sqrt{9}}{2\left(-3\right)}$$

Take the square root of 9.

$$x=\frac{-\left(-39\right)±3}{2\left(-3\right)}$$

The opposite of -39 is 39.

$$x=\frac{39±3}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{39±3}{-6}$$

Now solve the equation x=\frac{39±3}{-6} when ± is plus. Add 39 to 3.

$$x=\frac{42}{-6}$$

Divide 42 by -6.

$$x=-7$$

Now solve the equation x=\frac{39±3}{-6} when ± is minus. Subtract 3 from 39.

$$x=\frac{36}{-6}$$

Divide 36 by -6.

$$x=-6$$

The equation is now solved.

$$x=-7 x=-6$$

Steps for Completing the Square

Add 5x to both sides.

$$-3x^{2}-44x-97+5x=29$$

Combine -44x and 5x to get -39x.

$$-3x^{2}-39x-97=29$$

Add 97 to both sides.

$$-3x^{2}-39x=29+97$$

Add 29 and 97 to get 126.

$$-3x^{2}-39x=126$$

Divide both sides by -3.

$$\frac{-3x^{2}-39x}{-3}=\frac{126}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{39}{-3}\right)x=\frac{126}{-3}$$

Divide -39 by -3.

$$x^{2}+13x=\frac{126}{-3}$$

Divide 126 by -3.

$$x^{2}+13x=-42$$

Divide 13, the coefficient of the x term, by 2 to get \frac{13}{2}. Then add the square of \frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+13x+\left(\frac{13}{2}\right)^{2}=-42+\left(\frac{13}{2}\right)^{2}$$

Square \frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+13x+\frac{169}{4}=-42+\frac{169}{4}$$

Add -42 to \frac{169}{4}.

$$x^{2}+13x+\frac{169}{4}=\frac{1}{4}$$

Factor x^{2}+13x+\frac{169}{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}.

$$\left(x+\frac{13}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x+\frac{13}{2}=\frac{1}{2} x+\frac{13}{2}=-\frac{1}{2}$$

Subtract \frac{13}{2} from both sides of the equation.

$$x=-6 x=-7$$