Solve t^2-109t+900=0

$${ t }^{ 2 } -109t+900 = 0$$

Solve for t

$$t=9$$

$$t=100$$

Steps Using Factoring

To solve the equation, factor t^{2}-109t+900 using formula t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). To find a and b, set up a system to be solved.

$$a+b=-109 ab=900$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 900.

$$-1,-900 -2,-450 -3,-300 -4,-225 -5,-180 -6,-150 -9,-100 -10,-90 -12,-75 -15,-60 -18,-50 -20,-45 -25,-36 -30,-30$$

Calculate the sum for each pair.

$$-1-900=-901 -2-450=-452 -3-300=-303 -4-225=-229 -5-180=-185 -6-150=-156 -9-100=-109 -10-90=-100 -12-75=-87 -15-60=-75 -18-50=-68 -20-45=-65 -25-36=-61 -30-30=-60$$

The solution is the pair that gives sum -109.

$$a=-100 b=-9$$

Rewrite factored expression \left(t+a\right)\left(t+b\right) using the obtained values.

$$\left(t-100\right)\left(t-9\right)$$

To find equation solutions, solve t-100=0 and t-9=0.

$$t=100 t=9$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt+900. To find a and b, set up a system to be solved.

$$a+b=-109 ab=1\times 900=900$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 900.

$$-1,-900 -2,-450 -3,-300 -4,-225 -5,-180 -6,-150 -9,-100 -10,-90 -12,-75 -15,-60 -18,-50 -20,-45 -25,-36 -30,-30$$

Calculate the sum for each pair.

$$-1-900=-901 -2-450=-452 -3-300=-303 -4-225=-229 -5-180=-185 -6-150=-156 -9-100=-109 -10-90=-100 -12-75=-87 -15-60=-75 -18-50=-68 -20-45=-65 -25-36=-61 -30-30=-60$$

The solution is the pair that gives sum -109.

$$a=-100 b=-9$$

Rewrite t^{2}-109t+900 as \left(t^{2}-100t\right)+\left(-9t+900\right).

$$\left(t^{2}-100t\right)+\left(-9t+900\right)$$

Factor out t in the first and -9 in the second group.

$$t\left(t-100\right)-9\left(t-100\right)$$

Factor out common term t-100 by using distributive property.

$$\left(t-100\right)\left(t-9\right)$$

To find equation solutions, solve t-100=0 and t-9=0.

$$t=100 t=9$$

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.

$$t^{2}-109t+900=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -109 for b, and 900 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-109\right)±\sqrt{\left(-109\right)^{2}-4\times 900}}{2}$$

Square -109.

$$t=\frac{-\left(-109\right)±\sqrt{11881-4\times 900}}{2}$$

Multiply -4 times 900.

$$t=\frac{-\left(-109\right)±\sqrt{11881-3600}}{2}$$

Add 11881 to -3600.

$$t=\frac{-\left(-109\right)±\sqrt{8281}}{2}$$

Take the square root of 8281.

$$t=\frac{-\left(-109\right)±91}{2}$$

The opposite of -109 is 109.

$$t=\frac{109±91}{2}$$

Now solve the equation t=\frac{109±91}{2} when ± is plus. Add 109 to 91.

$$t=\frac{200}{2}$$

Divide 200 by 2.

$$t=100$$

Now solve the equation t=\frac{109±91}{2} when ± is minus. Subtract 91 from 109.

$$t=\frac{18}{2}$$

Divide 18 by 2.

$$t=9$$

The equation is now solved.

$$t=100 t=9$$

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.

$$t^{2}-109t+900=0$$

Subtract 900 from both sides of the equation.

$$t^{2}-109t+900-900=-900$$

Subtracting 900 from itself leaves 0.

$$t^{2}-109t=-900$$

Divide -109, the coefficient of the x term, by 2 to get -\frac{109}{2}. Then add the square of -\frac{109}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-109t+\left(-\frac{109}{2}\right)^{2}=-900+\left(-\frac{109}{2}\right)^{2}$$

Square -\frac{109}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-109t+\frac{11881}{4}=-900+\frac{11881}{4}$$

Add -900 to \frac{11881}{4}.

$$t^{2}-109t+\frac{11881}{4}=\frac{8281}{4}$$

Factor t^{2}-109t+\frac{11881}{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}.

$$\left(t-\frac{109}{2}\right)^{2}=\frac{8281}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{109}{2}\right)^{2}}=\sqrt{\frac{8281}{4}}$$

Simplify.

$$t-\frac{109}{2}=\frac{91}{2} t-\frac{109}{2}=-\frac{91}{2}$$

Add \frac{109}{2} to both sides of the equation.

$$t=100 t=9$$