Solve 10-5t^2=10t-5

$$10 - 5 t ^ { 2 } = 10 t - 5$$

Solve for t

$$t=-3$$

$$t=1$$

Steps Using Factoring By Grouping

Subtract 10t from both sides.

$$10-5t^{2}-10t=-5$$

Add 5 to both sides.

$$10-5t^{2}-10t+5=0$$

Add 10 and 5 to get 15.

$$15-5t^{2}-10t=0$$

Divide both sides by 5.

$$3-t^{2}-2t=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-t^{2}-2t+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -t^{2}+at+bt+3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-3$$

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

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

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

$$t\left(-t+1\right)+3\left(-t+1\right)$$

Factor out common term -t+1 by using distributive property.

$$\left(-t+1\right)\left(t+3\right)$$

To find equation solutions, solve -t+1=0 and t+3=0.

$$t=1 t=-3$$

Steps Using the Quadratic Formula

Subtract 10t from both sides.

$$10-5t^{2}-10t=-5$$

Add 5 to both sides.

$$10-5t^{2}-10t+5=0$$

Add 10 and 5 to get 15.

$$15-5t^{2}-10t=0$$

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.

$$-5t^{2}-10t+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -10 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-5\right)\times 15}}{2\left(-5\right)}$$

Square -10.

$$t=\frac{-\left(-10\right)±\sqrt{100-4\left(-5\right)\times 15}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-\left(-10\right)±\sqrt{100+20\times 15}}{2\left(-5\right)}$$

Multiply 20 times 15.

$$t=\frac{-\left(-10\right)±\sqrt{100+300}}{2\left(-5\right)}$$

Add 100 to 300.

$$t=\frac{-\left(-10\right)±\sqrt{400}}{2\left(-5\right)}$$

Take the square root of 400.

$$t=\frac{-\left(-10\right)±20}{2\left(-5\right)}$$

The opposite of -10 is 10.

$$t=\frac{10±20}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{10±20}{-10}$$

Now solve the equation t=\frac{10±20}{-10} when ± is plus. Add 10 to 20.

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

Divide 30 by -10.

$$t=-3$$

Now solve the equation t=\frac{10±20}{-10} when ± is minus. Subtract 20 from 10.

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

Divide -10 by -10.

$$t=1$$

The equation is now solved.

$$t=-3 t=1$$

Steps for Completing the Square

Subtract 10t from both sides.

$$10-5t^{2}-10t=-5$$

Subtract 10 from both sides.

$$-5t^{2}-10t=-5-10$$

Subtract 10 from -5 to get -15.

$$-5t^{2}-10t=-15$$

Divide both sides by -5.

$$\frac{-5t^{2}-10t}{-5}=-\frac{15}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$t^{2}+\left(-\frac{10}{-5}\right)t=-\frac{15}{-5}$$

Divide -10 by -5.

$$t^{2}+2t=-\frac{15}{-5}$$

Divide -15 by -5.

$$t^{2}+2t=3$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+2t+1^{2}=3+1^{2}$$

Square 1.

$$t^{2}+2t+1=3+1$$

Add 3 to 1.

$$t^{2}+2t+1=4$$

Factor t^{2}+2t+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$t+1=2 t+1=-2$$

Subtract 1 from both sides of the equation.

$$t=1 t=-3$$