Solve 3x^2=19x+14

$$3 x ^ { 2 } = 19 x + 14$$

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=7$$

Steps Using Factoring By Grouping

Subtract 19x from both sides.

$$3x^{2}-19x=14$$

Subtract 14 from both sides.

$$3x^{2}-19x-14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-14. To find a and b, set up a system to be solved.

$$a+b=-19 ab=3\left(-14\right)=-42$$

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 -19.

$$a=-21 b=2$$

Rewrite 3x^{2}-19x-14 as \left(3x^{2}-21x\right)+\left(2x-14\right).

$$\left(3x^{2}-21x\right)+\left(2x-14\right)$$

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

$$3x\left(x-7\right)+2\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(3x+2\right)$$

To find equation solutions, solve x-7=0 and 3x+2=0.

$$x=7 x=-\frac{2}{3}$$

Steps Using the Quadratic Formula

Subtract 19x from both sides.

$$3x^{2}-19x=14$$

Subtract 14 from both sides.

$$3x^{2}-19x-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -19 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 3\left(-14\right)}}{2\times 3}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 3\left(-14\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-19\right)±\sqrt{361-12\left(-14\right)}}{2\times 3}$$

Multiply -12 times -14.

$$x=\frac{-\left(-19\right)±\sqrt{361+168}}{2\times 3}$$

Add 361 to 168.

$$x=\frac{-\left(-19\right)±\sqrt{529}}{2\times 3}$$

Take the square root of 529.

$$x=\frac{-\left(-19\right)±23}{2\times 3}$$

The opposite of -19 is 19.

$$x=\frac{19±23}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{19±23}{6}$$

Now solve the equation x=\frac{19±23}{6} when ± is plus. Add 19 to 23.

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

Divide 42 by 6.

$$x=7$$

Now solve the equation x=\frac{19±23}{6} when ± is minus. Subtract 23 from 19.

$$x=-\frac{4}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=7 x=-\frac{2}{3}$$

Steps for Completing the Square

Subtract 19x from both sides.

$$3x^{2}-19x=14$$

Divide both sides by 3.

$$\frac{3x^{2}-19x}{3}=\frac{14}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{19}{3}x=\frac{14}{3}$$

Divide -\frac{19}{3}, the coefficient of the x term, by 2 to get -\frac{19}{6}. Then add the square of -\frac{19}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{19}{3}x+\left(-\frac{19}{6}\right)^{2}=\frac{14}{3}+\left(-\frac{19}{6}\right)^{2}$$

Square -\frac{19}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{19}{3}x+\frac{361}{36}=\frac{14}{3}+\frac{361}{36}$$

Add \frac{14}{3} to \frac{361}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{19}{3}x+\frac{361}{36}=\frac{529}{36}$$

Factor x^{2}-\frac{19}{3}x+\frac{361}{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}.

$$\left(x-\frac{19}{6}\right)^{2}=\frac{529}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{19}{6}\right)^{2}}=\sqrt{\frac{529}{36}}$$

Simplify.

$$x-\frac{19}{6}=\frac{23}{6} x-\frac{19}{6}=-\frac{23}{6}$$

Add \frac{19}{6} to both sides of the equation.

$$x=7 x=-\frac{2}{3}$$