Solve 25t-5t^2-30

$$25 t - 5 t ^ { 2 } - 30$$

Factor

$$5\left(2-t\right)\left(t-3\right)$$

Solution Steps

Factor out 5.

$$5\left(5t-t^{2}-6\right)$$

Consider 5t-t^{2}-6. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-t^{2}+5t-6$$

Factor the expression by grouping. First, the expression needs to be rewritten as -t^{2}+at+bt-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=-\left(-6\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=3 b=2$$

Rewrite -t^{2}+5t-6 as \left(-t^{2}+3t\right)+\left(2t-6\right).

$$\left(-t^{2}+3t\right)+\left(2t-6\right)$$

Factor out -t in the first and 2 in the second group.

$$-t\left(t-3\right)+2\left(t-3\right)$$

Factor out common term t-3 by using distributive property.

$$\left(t-3\right)\left(-t+2\right)$$

Rewrite the complete factored expression.

$$5\left(t-3\right)\left(-t+2\right)$$

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.

$$-5t^{2}+25t-30=0$$

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.

$$t=\frac{-25±\sqrt{25^{2}-4\left(-5\right)\left(-30\right)}}{2\left(-5\right)}$$

Square 25.

$$t=\frac{-25±\sqrt{625-4\left(-5\right)\left(-30\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-25±\sqrt{625+20\left(-30\right)}}{2\left(-5\right)}$$

Multiply 20 times -30.

$$t=\frac{-25±\sqrt{625-600}}{2\left(-5\right)}$$

Add 625 to -600.

$$t=\frac{-25±\sqrt{25}}{2\left(-5\right)}$$

Take the square root of 25.

$$t=\frac{-25±5}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-25±5}{-10}$$

Now solve the equation t=\frac{-25±5}{-10} when ± is plus. Add -25 to 5.

$$t=-\frac{20}{-10}$$

Divide -20 by -10.

$$t=2$$

Now solve the equation t=\frac{-25±5}{-10} when ± is minus. Subtract 5 from -25.

$$t=-\frac{30}{-10}$$

Divide -30 by -10.

$$t=3$$

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

$$-5t^{2}+25t-30=-5\left(t-2\right)\left(t-3\right)$$

Evaluate

$$5\left(2-t\right)\left(t-3\right)$$