Solve 3x^2+13x-1596=0

$$3 { x }^{ 2 } +13x-1596 = 0$$

Solve for x

$$x = -\frac{76}{3} = -25\frac{1}{3} \approx -25.333333333$$

$$x=21$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-1596. To find a and b, set up a system to be solved.

$$a+b=13 ab=3\left(-1596\right)=-4788$$

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. List all such integer pairs that give product -4788.

$$-1,4788 -2,2394 -3,1596 -4,1197 -6,798 -7,684 -9,532 -12,399 -14,342 -18,266 -19,252 -21,228 -28,171 -36,133 -38,126 -42,114 -57,84 -63,76$$

Calculate the sum for each pair.

$$-1+4788=4787 -2+2394=2392 -3+1596=1593 -4+1197=1193 -6+798=792 -7+684=677 -9+532=523 -12+399=387 -14+342=328 -18+266=248 -19+252=233 -21+228=207 -28+171=143 -36+133=97 -38+126=88 -42+114=72 -57+84=27 -63+76=13$$

The solution is the pair that gives sum 13.

$$a=-63 b=76$$

Rewrite 3x^{2}+13x-1596 as \left(3x^{2}-63x\right)+\left(76x-1596\right).

$$\left(3x^{2}-63x\right)+\left(76x-1596\right)$$

Factor out 3x in the first and 76 in the second group.

$$3x\left(x-21\right)+76\left(x-21\right)$$

Factor out common term x-21 by using distributive property.

$$\left(x-21\right)\left(3x+76\right)$$

To find equation solutions, solve x-21=0 and 3x+76=0.

$$x=21 x=-\frac{76}{3}$$

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.

$$3x^{2}+13x-1596=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 13 for b, and -1596 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\times 3\left(-1596\right)}}{2\times 3}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 3\left(-1596\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-13±\sqrt{169-12\left(-1596\right)}}{2\times 3}$$

Multiply -12 times -1596.

$$x=\frac{-13±\sqrt{169+19152}}{2\times 3}$$

Add 169 to 19152.

$$x=\frac{-13±\sqrt{19321}}{2\times 3}$$

Take the square root of 19321.

$$x=\frac{-13±139}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-13±139}{6}$$

Now solve the equation x=\frac{-13±139}{6} when ± is plus. Add -13 to 139.

$$x=\frac{126}{6}$$

Divide 126 by 6.

$$x=21$$

Now solve the equation x=\frac{-13±139}{6} when ± is minus. Subtract 139 from -13.

$$x=-\frac{152}{6}$$

Reduce the fraction \frac{-152}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{76}{3}$$

The equation is now solved.

$$x=21 x=-\frac{76}{3}$$

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.

$$3x^{2}+13x-1596=0$$

Add 1596 to both sides of the equation.

$$3x^{2}+13x-1596-\left(-1596\right)=-\left(-1596\right)$$

Subtracting -1596 from itself leaves 0.

$$3x^{2}+13x=-\left(-1596\right)$$

Subtract -1596 from 0.

$$3x^{2}+13x=1596$$

Divide both sides by 3.

$$\frac{3x^{2}+13x}{3}=\frac{1596}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{13}{3}x=\frac{1596}{3}$$

Divide 1596 by 3.

$$x^{2}+\frac{13}{3}x=532$$

Divide \frac{13}{3}, the coefficient of the x term, by 2 to get \frac{13}{6}. Then add the square of \frac{13}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{3}x+\left(\frac{13}{6}\right)^{2}=532+\left(\frac{13}{6}\right)^{2}$$

Square \frac{13}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{3}x+\frac{169}{36}=532+\frac{169}{36}$$

Add 532 to \frac{169}{36}.

$$x^{2}+\frac{13}{3}x+\frac{169}{36}=\frac{19321}{36}$$

Factor x^{2}+\frac{13}{3}x+\frac{169}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{13}{6}\right)^{2}=\frac{19321}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{6}\right)^{2}}=\sqrt{\frac{19321}{36}}$$

Simplify.

$$x+\frac{13}{6}=\frac{139}{6} x+\frac{13}{6}=-\frac{139}{6}$$

Subtract \frac{13}{6} from both sides of the equation.

$$x=21 x=-\frac{76}{3}$$