Solve 150=1/2b(b-5

$$150 = 1 / 2 b ( b - 5$$

Solve for b

$$b=-15$$

$$b=20$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{2}b by b-5.

$$150=\frac{1}{2}bb+\frac{1}{2}b\left(-5\right)$$

Multiply b and b to get b^{2}.

$$150=\frac{1}{2}b^{2}+\frac{1}{2}b\left(-5\right)$$

Multiply \frac{1}{2} and -5 to get \frac{-5}{2}.

$$150=\frac{1}{2}b^{2}+\frac{-5}{2}b$$

Fraction \frac{-5}{2} can be rewritten as -\frac{5}{2} by extracting the negative sign.

$$150=\frac{1}{2}b^{2}-\frac{5}{2}b$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}b^{2}-\frac{5}{2}b=150$$

Subtract 150 from both sides.

$$\frac{1}{2}b^{2}-\frac{5}{2}b-150=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -\frac{5}{2} for b, and -150 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\left(-\frac{5}{2}\right)^{2}-4\times \frac{1}{2}\left(-150\right)}}{2\times \frac{1}{2}}$$

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

$$b=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-4\times \frac{1}{2}\left(-150\right)}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$b=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-2\left(-150\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -150.

$$b=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}+300}}{2\times \frac{1}{2}}$$

Add \frac{25}{4} to 300.

$$b=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{1225}{4}}}{2\times \frac{1}{2}}$$

Take the square root of \frac{1225}{4}.

$$b=\frac{-\left(-\frac{5}{2}\right)±\frac{35}{2}}{2\times \frac{1}{2}}$$

The opposite of -\frac{5}{2} is \frac{5}{2}.

$$b=\frac{\frac{5}{2}±\frac{35}{2}}{2\times \frac{1}{2}}$$

Multiply 2 times \frac{1}{2}.

$$b=\frac{\frac{5}{2}±\frac{35}{2}}{1}$$

Now solve the equation b=\frac{\frac{5}{2}±\frac{35}{2}}{1} when ± is plus. Add \frac{5}{2} to \frac{35}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b=\frac{20}{1}$$

Divide 20 by 1.

$$b=20$$

Now solve the equation b=\frac{\frac{5}{2}±\frac{35}{2}}{1} when ± is minus. Subtract \frac{35}{2} from \frac{5}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$b=-\frac{15}{1}$$

Divide -15 by 1.

$$b=-15$$

The equation is now solved.

$$b=20 b=-15$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{2}b by b-5.

$$150=\frac{1}{2}bb+\frac{1}{2}b\left(-5\right)$$

Multiply b and b to get b^{2}.

$$150=\frac{1}{2}b^{2}+\frac{1}{2}b\left(-5\right)$$

Multiply \frac{1}{2} and -5 to get \frac{-5}{2}.

$$150=\frac{1}{2}b^{2}+\frac{-5}{2}b$$

Fraction \frac{-5}{2} can be rewritten as -\frac{5}{2} by extracting the negative sign.

$$150=\frac{1}{2}b^{2}-\frac{5}{2}b$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}b^{2}-\frac{5}{2}b=150$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}b^{2}-\frac{5}{2}b}{\frac{1}{2}}=\frac{150}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$b^{2}+\left(-\frac{\frac{5}{2}}{\frac{1}{2}}\right)b=\frac{150}{\frac{1}{2}}$$

Divide -\frac{5}{2} by \frac{1}{2} by multiplying -\frac{5}{2} by the reciprocal of \frac{1}{2}.

$$b^{2}-5b=\frac{150}{\frac{1}{2}}$$

Divide 150 by \frac{1}{2} by multiplying 150 by the reciprocal of \frac{1}{2}.

$$b^{2}-5b=300$$

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.

$$b^{2}-5b+\left(-\frac{5}{2}\right)^{2}=300+\left(-\frac{5}{2}\right)^{2}$$

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

$$b^{2}-5b+\frac{25}{4}=300+\frac{25}{4}$$

Add 300 to \frac{25}{4}.

$$b^{2}-5b+\frac{25}{4}=\frac{1225}{4}$$

Factor b^{2}-5b+\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}.

$$\left(b-\frac{5}{2}\right)^{2}=\frac{1225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1225}{4}}$$

Simplify.

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

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

$$b=20 b=-15$$