Solve 5x^2-23x+24=2x^2-8x+6

$$5 x ^ { 2 } - 23 x + 24 = 2 x ^ { 2 } - 8 x + 6$$

Solve for x

$$x=2$$

$$x=3$$

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

$$5x^{2}-23x+24-2x^{2}=-8x+6$$

Combine 5x^{2} and -2x^{2} to get 3x^{2}.

$$3x^{2}-23x+24=-8x+6$$

Add 8x to both sides.

$$3x^{2}-23x+24+8x=6$$

Combine -23x and 8x to get -15x.

$$3x^{2}-15x+24=6$$

Subtract 6 from both sides.

$$3x^{2}-15x+24-6=0$$

Subtract 6 from 24 to get 18.

$$3x^{2}-15x+18=0$$

Divide both sides by 3.

$$x^{2}-5x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 6=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite x^{2}-5x+6 as \left(x^{2}-3x\right)+\left(-2x+6\right).

$$\left(x^{2}-3x\right)+\left(-2x+6\right)$$

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

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

Factor out common term x-3 by using distributive property.

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

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

$$x=3 x=2$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$5x^{2}-23x+24-2x^{2}=-8x+6$$

Combine 5x^{2} and -2x^{2} to get 3x^{2}.

$$3x^{2}-23x+24=-8x+6$$

Add 8x to both sides.

$$3x^{2}-23x+24+8x=6$$

Combine -23x and 8x to get -15x.

$$3x^{2}-15x+24=6$$

Subtract 6 from both sides.

$$3x^{2}-15x+24-6=0$$

Subtract 6 from 24 to get 18.

$$3x^{2}-15x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -15 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 3\times 18}}{2\times 3}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 3\times 18}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-15\right)±\sqrt{225-12\times 18}}{2\times 3}$$

Multiply -12 times 18.

$$x=\frac{-\left(-15\right)±\sqrt{225-216}}{2\times 3}$$

Add 225 to -216.

$$x=\frac{-\left(-15\right)±\sqrt{9}}{2\times 3}$$

Take the square root of 9.

$$x=\frac{-\left(-15\right)±3}{2\times 3}$$

The opposite of -15 is 15.

$$x=\frac{15±3}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{15±3}{6}$$

Now solve the equation x=\frac{15±3}{6} when ± is plus. Add 15 to 3.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{15±3}{6} when ± is minus. Subtract 3 from 15.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

The equation is now solved.

$$x=3 x=2$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$5x^{2}-23x+24-2x^{2}=-8x+6$$

Combine 5x^{2} and -2x^{2} to get 3x^{2}.

$$3x^{2}-23x+24=-8x+6$$

Add 8x to both sides.

$$3x^{2}-23x+24+8x=6$$

Combine -23x and 8x to get -15x.

$$3x^{2}-15x+24=6$$

Subtract 24 from both sides.

$$3x^{2}-15x=6-24$$

Subtract 24 from 6 to get -18.

$$3x^{2}-15x=-18$$

Divide both sides by 3.

$$\frac{3x^{2}-15x}{3}=-\frac{18}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{15}{3}\right)x=-\frac{18}{3}$$

Divide -15 by 3.

$$x^{2}-5x=-\frac{18}{3}$$

Divide -18 by 3.

$$x^{2}-5x=-6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=3 x=2$$