Solve x33-x+2/2=150

$$x \frac{ 33-x+2 }{ 2 } = 150$$

Solve for x

$$x=15$$

$$x=20$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$x\left(33-x+2\right)=300$$

Add 33 and 2 to get 35.

$$x\left(35-x\right)=300$$

Use the distributive property to multiply x by 35-x.

$$35x-x^{2}=300$$

Subtract 300 from both sides.

$$35x-x^{2}-300=0$$

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.

$$-x^{2}+35x-300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 35 for b, and -300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-35±\sqrt{35^{2}-4\left(-1\right)\left(-300\right)}}{2\left(-1\right)}$$

Square 35.

$$x=\frac{-35±\sqrt{1225-4\left(-1\right)\left(-300\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-35±\sqrt{1225+4\left(-300\right)}}{2\left(-1\right)}$$

Multiply 4 times -300.

$$x=\frac{-35±\sqrt{1225-1200}}{2\left(-1\right)}$$

Add 1225 to -1200.

$$x=\frac{-35±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$x=\frac{-35±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-35±5}{-2}$$

Now solve the equation x=\frac{-35±5}{-2} when ± is plus. Add -35 to 5.

$$x=-\frac{30}{-2}$$

Divide -30 by -2.

$$x=15$$

Now solve the equation x=\frac{-35±5}{-2} when ± is minus. Subtract 5 from -35.

$$x=-\frac{40}{-2}$$

Divide -40 by -2.

$$x=20$$

The equation is now solved.

$$x=15 x=20$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$x\left(33-x+2\right)=300$$

Add 33 and 2 to get 35.

$$x\left(35-x\right)=300$$

Use the distributive property to multiply x by 35-x.

$$35x-x^{2}=300$$

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.

$$-x^{2}+35x=300$$

Divide both sides by -1.

$$\frac{-x^{2}+35x}{-1}=\frac{300}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{35}{-1}x=\frac{300}{-1}$$

Divide 35 by -1.

$$x^{2}-35x=\frac{300}{-1}$$

Divide 300 by -1.

$$x^{2}-35x=-300$$

Divide -35, the coefficient of the x term, by 2 to get -\frac{35}{2}. Then add the square of -\frac{35}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-35x+\left(-\frac{35}{2}\right)^{2}=-300+\left(-\frac{35}{2}\right)^{2}$$

Square -\frac{35}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-35x+\frac{1225}{4}=-300+\frac{1225}{4}$$

Add -300 to \frac{1225}{4}.

$$x^{2}-35x+\frac{1225}{4}=\frac{25}{4}$$

Factor x^{2}-35x+\frac{1225}{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}.

$$\left(x-\frac{35}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{35}{2}=\frac{5}{2} x-\frac{35}{2}=-\frac{5}{2}$$

Add \frac{35}{2} to both sides of the equation.

$$x=20 x=15$$