Solve (x+0)(35-2x)=150

$$(x+0)(35-2x)=150$$

Solve for x

$$x=10$$

$$x = \frac{15}{2} = 7\frac{1}{2} = 7.5$$

Steps Using the Quadratic Formula

Anything plus zero gives itself.

$$x\left(35-2x\right)=150$$

Use the distributive property to multiply x by 35-2x.

$$35x-2x^{2}=150$$

Subtract 150 from both sides.

$$35x-2x^{2}-150=0$$

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.

$$-2x^{2}+35x-150=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 35 for b, and -150 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-35±\sqrt{35^{2}-4\left(-2\right)\left(-150\right)}}{2\left(-2\right)}$$

Square 35.

$$x=\frac{-35±\sqrt{1225-4\left(-2\right)\left(-150\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-35±\sqrt{1225+8\left(-150\right)}}{2\left(-2\right)}$$

Multiply 8 times -150.

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

Add 1225 to -1200.

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

Take the square root of 25.

$$x=\frac{-35±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-35±5}{-4}$$

Now solve the equation x=\frac{-35±5}{-4} when ± is plus. Add -35 to 5.

$$x=-\frac{30}{-4}$$

Reduce the fraction \frac{-30}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{15}{2}$$

Now solve the equation x=\frac{-35±5}{-4} when ± is minus. Subtract 5 from -35.

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

Divide -40 by -4.

$$x=10$$

The equation is now solved.

$$x=\frac{15}{2} x=10$$

Steps for Completing the Square

Anything plus zero gives itself.

$$x\left(35-2x\right)=150$$

Use the distributive property to multiply x by 35-2x.

$$35x-2x^{2}=150$$

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.

$$-2x^{2}+35x=150$$

Divide both sides by -2.

$$\frac{-2x^{2}+35x}{-2}=\frac{150}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{35}{-2}x=\frac{150}{-2}$$

Divide 35 by -2.

$$x^{2}-\frac{35}{2}x=\frac{150}{-2}$$

Divide 150 by -2.

$$x^{2}-\frac{35}{2}x=-75$$

Divide -\frac{35}{2}, the coefficient of the x term, by 2 to get -\frac{35}{4}. Then add the square of -\frac{35}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{35}{2}x+\left(-\frac{35}{4}\right)^{2}=-75+\left(-\frac{35}{4}\right)^{2}$$

Square -\frac{35}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{35}{2}x+\frac{1225}{16}=-75+\frac{1225}{16}$$

Add -75 to \frac{1225}{16}.

$$x^{2}-\frac{35}{2}x+\frac{1225}{16}=\frac{25}{16}$$

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

$$\left(x-\frac{35}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{35}{4}=\frac{5}{4} x-\frac{35}{4}=-\frac{5}{4}$$

Add \frac{35}{4} to both sides of the equation.

$$x=10 x=\frac{15}{2}$$