Solve 130t-5t^2

$$130 t - 5 t ^ { 2 }$$

Factor

$$5t\left(26-t\right)$$

Solution Steps

Factor out 5.

$$5\left(26t-t^{2}\right)$$

Consider 26t-t^{2}. Factor out t.

$$t\left(26-t\right)$$

Rewrite the complete factored expression.

$$5t\left(-t+26\right)$$

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.

$$-5t^{2}+130t=0$$

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.

$$t=\frac{-130±\sqrt{130^{2}}}{2\left(-5\right)}$$

Take the square root of 130^{2}.

$$t=\frac{-130±130}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-130±130}{-10}$$

Now solve the equation t=\frac{-130±130}{-10} when ± is plus. Add -130 to 130.

$$t=\frac{0}{-10}$$

Divide 0 by -10.

$$t=0$$

Now solve the equation t=\frac{-130±130}{-10} when ± is minus. Subtract 130 from -130.

$$t=-\frac{260}{-10}$$

Divide -260 by -10.

$$t=26$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and 26 for x_{2}.

$$-5t^{2}+130t=-5t\left(t-26\right)$$

Evaluate

$$5t\left(26-t\right)$$