Solve 2x^2-24x+25x-300=0

$$2 x ^ { 2 } - 24 x + 25 x - 300 = 0$$

Solve for x

$$x = -\frac{25}{2} = -12\frac{1}{2} = -12.5$$

$$x=12$$

Steps Using Factoring By Grouping

Combine -24x and 25x to get x.

$$2x^{2}+x-300=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-300. To find a and b, set up a system to be solved.

$$a+b=1 ab=2\left(-300\right)=-600$$

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 -600.

$$-1,600 -2,300 -3,200 -4,150 -5,120 -6,100 -8,75 -10,60 -12,50 -15,40 -20,30 -24,25$$

Calculate the sum for each pair.

$$-1+600=599 -2+300=298 -3+200=197 -4+150=146 -5+120=115 -6+100=94 -8+75=67 -10+60=50 -12+50=38 -15+40=25 -20+30=10 -24+25=1$$

The solution is the pair that gives sum 1.

$$a=-24 b=25$$

Rewrite 2x^{2}+x-300 as \left(2x^{2}-24x\right)+\left(25x-300\right).

$$\left(2x^{2}-24x\right)+\left(25x-300\right)$$

Factor out 2x in the first and 25 in the second group.

$$2x\left(x-12\right)+25\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(2x+25\right)$$

To find equation solutions, solve x-12=0 and 2x+25=0.

$$x=12 x=-\frac{25}{2}$$

Steps Using the Quadratic Formula

Combine -24x and 25x to get x.

$$2x^{2}+x-300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 1 for b, and -300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-300\right)}}{2\times 2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-300\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-300\right)}}{2\times 2}$$

Multiply -8 times -300.

$$x=\frac{-1±\sqrt{1+2400}}{2\times 2}$$

Add 1 to 2400.

$$x=\frac{-1±\sqrt{2401}}{2\times 2}$$

Take the square root of 2401.

$$x=\frac{-1±49}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±49}{4}$$

Now solve the equation x=\frac{-1±49}{4} when ± is plus. Add -1 to 49.

$$x=\frac{48}{4}$$

Divide 48 by 4.

$$x=12$$

Now solve the equation x=\frac{-1±49}{4} when ± is minus. Subtract 49 from -1.

$$x=-\frac{50}{4}$$

Reduce the fraction \frac{-50}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{25}{2}$$

The equation is now solved.

$$x=12 x=-\frac{25}{2}$$

Steps for Completing the Square

Combine -24x and 25x to get x.

$$2x^{2}+x-300=0$$

Add 300 to both sides. Anything plus zero gives itself.

$$2x^{2}+x=300$$

Divide both sides by 2.

$$\frac{2x^{2}+x}{2}=\frac{300}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{1}{2}x=\frac{300}{2}$$

Divide 300 by 2.

$$x^{2}+\frac{1}{2}x=150$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=150+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=150+\frac{1}{16}$$

Add 150 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{2401}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{2401}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{2401}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{49}{4} x+\frac{1}{4}=-\frac{49}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=12 x=-\frac{25}{2}$$