Solve 2x^2=19x-24

$$2 x ^ { 2 } = 19 x - 24$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=8$$

Steps Using Factoring By Grouping

Subtract 19x from both sides.

$$2x^{2}-19x=-24$$

Add 24 to both sides.

$$2x^{2}-19x+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=-19 ab=2\times 24=48$$

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 48.

$$-1,-48 -2,-24 -3,-16 -4,-12 -6,-8$$

Calculate the sum for each pair.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -19.

$$a=-16 b=-3$$

Rewrite 2x^{2}-19x+24 as \left(2x^{2}-16x\right)+\left(-3x+24\right).

$$\left(2x^{2}-16x\right)+\left(-3x+24\right)$$

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

$$2x\left(x-8\right)-3\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(2x-3\right)$$

To find equation solutions, solve x-8=0 and 2x-3=0.

$$x=8 x=\frac{3}{2}$$

Steps Using the Quadratic Formula

Subtract 19x from both sides.

$$2x^{2}-19x=-24$$

Add 24 to both sides.

$$2x^{2}-19x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -19 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 2\times 24}}{2\times 2}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 2\times 24}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-19\right)±\sqrt{361-8\times 24}}{2\times 2}$$

Multiply -8 times 24.

$$x=\frac{-\left(-19\right)±\sqrt{361-192}}{2\times 2}$$

Add 361 to -192.

$$x=\frac{-\left(-19\right)±\sqrt{169}}{2\times 2}$$

Take the square root of 169.

$$x=\frac{-\left(-19\right)±13}{2\times 2}$$

The opposite of -19 is 19.

$$x=\frac{19±13}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{19±13}{4}$$

Now solve the equation x=\frac{19±13}{4} when ± is plus. Add 19 to 13.

$$x=\frac{32}{4}$$

Divide 32 by 4.

$$x=8$$

Now solve the equation x=\frac{19±13}{4} when ± is minus. Subtract 13 from 19.

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

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

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

The equation is now solved.

$$x=8 x=\frac{3}{2}$$

Steps for Completing the Square

Subtract 19x from both sides.

$$2x^{2}-19x=-24$$

Divide both sides by 2.

$$\frac{2x^{2}-19x}{2}=-\frac{24}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{19}{2}x=-\frac{24}{2}$$

Divide -24 by 2.

$$x^{2}-\frac{19}{2}x=-12$$

Divide -\frac{19}{2}, the coefficient of the x term, by 2 to get -\frac{19}{4}. Then add the square of -\frac{19}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{19}{2}x+\left(-\frac{19}{4}\right)^{2}=-12+\left(-\frac{19}{4}\right)^{2}$$

Square -\frac{19}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{19}{2}x+\frac{361}{16}=-12+\frac{361}{16}$$

Add -12 to \frac{361}{16}.

$$x^{2}-\frac{19}{2}x+\frac{361}{16}=\frac{169}{16}$$

Factor x^{2}-\frac{19}{2}x+\frac{361}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{19}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{19}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x-\frac{19}{4}=\frac{13}{4} x-\frac{19}{4}=-\frac{13}{4}$$

Add \frac{19}{4} to both sides of the equation.

$$x=8 x=\frac{3}{2}$$