Solve 840=130*t-5t^2

$$840 = 130 \cdot t - 5 t ^ { 2 }$$

Solve for t

$$t=12$$

$$t=14$$

Steps Using the Quadratic Formula

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

$$130t-5t^{2}=840$$

Subtract 840 from both sides.

$$130t-5t^{2}-840=0$$

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.

$$-5t^{2}+130t-840=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 130 for b, and -840 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-130±\sqrt{130^{2}-4\left(-5\right)\left(-840\right)}}{2\left(-5\right)}$$

Square 130.

$$t=\frac{-130±\sqrt{16900-4\left(-5\right)\left(-840\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-130±\sqrt{16900+20\left(-840\right)}}{2\left(-5\right)}$$

Multiply 20 times -840.

$$t=\frac{-130±\sqrt{16900-16800}}{2\left(-5\right)}$$

Add 16900 to -16800.

$$t=\frac{-130±\sqrt{100}}{2\left(-5\right)}$$

Take the square root of 100.

$$t=\frac{-130±10}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-130±10}{-10}$$

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

$$t=-\frac{120}{-10}$$

Divide -120 by -10.

$$t=12$$

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

$$t=-\frac{140}{-10}$$

Divide -140 by -10.

$$t=14$$

The equation is now solved.

$$t=12 t=14$$

Steps for Completing the Square

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

$$130t-5t^{2}=840$$

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.

$$-5t^{2}+130t=840$$

Divide both sides by -5.

$$\frac{-5t^{2}+130t}{-5}=\frac{840}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$t^{2}+\frac{130}{-5}t=\frac{840}{-5}$$

Divide 130 by -5.

$$t^{2}-26t=\frac{840}{-5}$$

Divide 840 by -5.

$$t^{2}-26t=-168$$

Divide -26, the coefficient of the x term, by 2 to get -13. Then add the square of -13 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-26t+\left(-13\right)^{2}=-168+\left(-13\right)^{2}$$

Square -13.

$$t^{2}-26t+169=-168+169$$

Add -168 to 169.

$$t^{2}-26t+169=1$$

Factor t^{2}-26t+169. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-13\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-13\right)^{2}}=\sqrt{1}$$

Simplify.

$$t-13=1 t-13=-1$$

Add 13 to both sides of the equation.

$$t=14 t=12$$