Solve t^2-28t+75

$$t ^ { 2 } - 28 t + 75$$

Factor

$$\left(t-25\right)\left(t-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt+75. To find a and b, set up a system to be solved.

$$a+b=-28 ab=1\times 75=75$$

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 75.

$$-1,-75 -3,-25 -5,-15$$

Calculate the sum for each pair.

$$-1-75=-76 -3-25=-28 -5-15=-20$$

The solution is the pair that gives sum -28.

$$a=-25 b=-3$$

Rewrite t^{2}-28t+75 as \left(t^{2}-25t\right)+\left(-3t+75\right).

$$\left(t^{2}-25t\right)+\left(-3t+75\right)$$

Factor out t in the first and -3 in the second group.

$$t\left(t-25\right)-3\left(t-25\right)$$

Factor out common term t-25 by using distributive property.

$$\left(t-25\right)\left(t-3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$t^{2}-28t+75=0$$

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.

$$t=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 75}}{2}$$

Square -28.

$$t=\frac{-\left(-28\right)±\sqrt{784-4\times 75}}{2}$$

Multiply -4 times 75.

$$t=\frac{-\left(-28\right)±\sqrt{784-300}}{2}$$

Add 784 to -300.

$$t=\frac{-\left(-28\right)±\sqrt{484}}{2}$$

Take the square root of 484.

$$t=\frac{-\left(-28\right)±22}{2}$$

The opposite of -28 is 28.

$$t=\frac{28±22}{2}$$

Now solve the equation t=\frac{28±22}{2} when ± is plus. Add 28 to 22.

$$t=\frac{50}{2}$$

Divide 50 by 2.

$$t=25$$

Now solve the equation t=\frac{28±22}{2} when ± is minus. Subtract 22 from 28.

$$t=\frac{6}{2}$$

Divide 6 by 2.

$$t=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 25 for x_{1} and 3 for x_{2}.

$$t^{2}-28t+75=\left(t-25\right)\left(t-3\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -28x +75 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 28 rs = 75$$

Two numbers r and s sum up to 28 exactly when the average of the two numbers is \frac{1}{2}*28 = 14. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 14 - u s = 14 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 75

$$(14 - u) (14 + u) = 75$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$196 - u^2 = 75$$

Simplify the expression by subtracting 196 on both sides

$$-u^2 = 75-196 = -121$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 121 u = \pm\sqrt{121} = \pm 11$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =14 - 11 = 3 s = 14 + 11 = 25$$

Evaluate

$$\left(t-25\right)\left(t-3\right)$$