Solve -12u-9=4u^2

$$- 12 u - 9 = 4 u ^ { 2 }$$

Solve for u

$$u = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps Using Factoring By Grouping

Subtract 4u^{2} from both sides.

$$-12u-9-4u^{2}=0$$

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

$$-4u^{2}-12u-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4u^{2}+au+bu-9. To find a and b, set up a system to be solved.

$$a+b=-12 ab=-4\left(-9\right)=36$$

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 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -12.

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

Rewrite -4u^{2}-12u-9 as \left(-4u^{2}-6u\right)+\left(-6u-9\right).

$$\left(-4u^{2}-6u\right)+\left(-6u-9\right)$$

Factor out 2u in the first and 3 in the second group.

$$2u\left(-2u-3\right)+3\left(-2u-3\right)$$

Factor out common term -2u-3 by using distributive property.

$$\left(-2u-3\right)\left(2u+3\right)$$

To find equation solutions, solve -2u-3=0 and 2u+3=0.

$$u=-\frac{3}{2} u=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Subtract 4u^{2} from both sides.

$$-12u-9-4u^{2}=0$$

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.

$$-4u^{2}-12u-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -12 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$u=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-4\right)\left(-9\right)}}{2\left(-4\right)}$$

Square -12.

$$u=\frac{-\left(-12\right)±\sqrt{144-4\left(-4\right)\left(-9\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$u=\frac{-\left(-12\right)±\sqrt{144+16\left(-9\right)}}{2\left(-4\right)}$$

Multiply 16 times -9.

$$u=\frac{-\left(-12\right)±\sqrt{144-144}}{2\left(-4\right)}$$

Add 144 to -144.

$$u=\frac{-\left(-12\right)±\sqrt{0}}{2\left(-4\right)}$$

Take the square root of 0.

$$u=-\frac{-12}{2\left(-4\right)}$$

The opposite of -12 is 12.

$$u=\frac{12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$u=\frac{12}{-8}$$

Reduce the fraction \frac{12}{-8} to lowest terms by extracting and canceling out 4.

$$u=-\frac{3}{2}$$

Steps for Completing the Square

Subtract 4u^{2} from both sides.

$$-12u-9-4u^{2}=0$$

Add 9 to both sides. Anything plus zero gives itself.

$$-12u-4u^{2}=9$$

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.

$$-4u^{2}-12u=9$$

Divide both sides by -4.

$$\frac{-4u^{2}-12u}{-4}=\frac{9}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$u^{2}+\left(-\frac{12}{-4}\right)u=\frac{9}{-4}$$

Divide -12 by -4.

$$u^{2}+3u=\frac{9}{-4}$$

Divide 9 by -4.

$$u^{2}+3u=-\frac{9}{4}$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$u^{2}+3u+\left(\frac{3}{2}\right)^{2}=-\frac{9}{4}+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$u^{2}+3u+\frac{9}{4}=\frac{-9+9}{4}$$

Add -\frac{9}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$u^{2}+3u+\frac{9}{4}=0$$

Factor u^{2}+3u+\frac{9}{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}.

$$\left(u+\frac{3}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(u+\frac{3}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$u+\frac{3}{2}=0 u+\frac{3}{2}=0$$

Subtract \frac{3}{2} from both sides of the equation.

$$u=-\frac{3}{2} u=-\frac{3}{2}$$

The equation is now solved. Solutions are the same.

$$u=-\frac{3}{2}$$