Solve 50x-2x^2-300=0

$$50 x - 2 x ^ { 2 } - 300 = 0$$

Solve for x

$$x=10$$

$$x=15$$

Steps Using Factoring By Grouping

Divide both sides by 2.

$$25x-x^{2}-150=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+25x-150=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-150. To find a and b, set up a system to be solved.

$$a+b=25 ab=-\left(-150\right)=150$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 150.

$$1,150 2,75 3,50 5,30 6,25 10,15$$

Calculate the sum for each pair.

$$1+150=151 2+75=77 3+50=53 5+30=35 6+25=31 10+15=25$$

The solution is the pair that gives sum 25.

$$a=15 b=10$$

Rewrite -x^{2}+25x-150 as \left(-x^{2}+15x\right)+\left(10x-150\right).

$$\left(-x^{2}+15x\right)+\left(10x-150\right)$$

Factor out -x in the first and 10 in the second group.

$$-x\left(x-15\right)+10\left(x-15\right)$$

Factor out common term x-15 by using distributive property.

$$\left(x-15\right)\left(-x+10\right)$$

To find equation solutions, solve x-15=0 and -x+10=0.

$$x=15 x=10$$

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.

$$-2x^{2}+50x-300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 50 for b, and -300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-50±\sqrt{50^{2}-4\left(-2\right)\left(-300\right)}}{2\left(-2\right)}$$

Square 50.

$$x=\frac{-50±\sqrt{2500-4\left(-2\right)\left(-300\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-50±\sqrt{2500+8\left(-300\right)}}{2\left(-2\right)}$$

Multiply 8 times -300.

$$x=\frac{-50±\sqrt{2500-2400}}{2\left(-2\right)}$$

Add 2500 to -2400.

$$x=\frac{-50±\sqrt{100}}{2\left(-2\right)}$$

Take the square root of 100.

$$x=\frac{-50±10}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-50±10}{-4}$$

Now solve the equation x=\frac{-50±10}{-4} when ± is plus. Add -50 to 10.

$$x=-\frac{40}{-4}$$

Divide -40 by -4.

$$x=10$$

Now solve the equation x=\frac{-50±10}{-4} when ± is minus. Subtract 10 from -50.

$$x=-\frac{60}{-4}$$

Divide -60 by -4.

$$x=15$$

The equation is now solved.

$$x=10 x=15$$

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.

$$-2x^{2}+50x-300=0$$

Add 300 to both sides of the equation.

$$-2x^{2}+50x-300-\left(-300\right)=-\left(-300\right)$$

Subtracting -300 from itself leaves 0.

$$-2x^{2}+50x=-\left(-300\right)$$

Subtract -300 from 0.

$$-2x^{2}+50x=300$$

Divide both sides by -2.

$$\frac{-2x^{2}+50x}{-2}=\frac{300}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{50}{-2}x=\frac{300}{-2}$$

Divide 50 by -2.

$$x^{2}-25x=\frac{300}{-2}$$

Divide 300 by -2.

$$x^{2}-25x=-150$$

Divide -25, the coefficient of the x term, by 2 to get -\frac{25}{2}. Then add the square of -\frac{25}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-25x+\left(-\frac{25}{2}\right)^{2}=-150+\left(-\frac{25}{2}\right)^{2}$$

Square -\frac{25}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-25x+\frac{625}{4}=-150+\frac{625}{4}$$

Add -150 to \frac{625}{4}.

$$x^{2}-25x+\frac{625}{4}=\frac{25}{4}$$

Factor x^{2}-25x+\frac{625}{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}.

$$\left(x-\frac{25}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{25}{2}=\frac{5}{2} x-\frac{25}{2}=-\frac{5}{2}$$

Add \frac{25}{2} to both sides of the equation.

$$x=15 x=10$$