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

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

Solve for t

$$t=2$$

Steps Using the Quadratic Formula

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.

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

Subtract 10 from both sides of the equation.

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

Subtracting 10 from itself leaves 0.

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

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

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

Square 10.

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

Multiply -4 times -\frac{5}{2}.

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

Multiply 10 times -10.

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

Add 100 to -100.

$$t=\frac{-10±\sqrt{0}}{2\left(-\frac{5}{2}\right)}$$

Take the square root of 0.

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

Multiply 2 times -\frac{5}{2}.

$$t=-\frac{10}{-5}$$

Divide -10 by -5.

$$t=2$$

Steps for Completing the Square

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.

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

Divide both sides of the equation by -\frac{5}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by -\frac{5}{2} undoes the multiplication by -\frac{5}{2}.

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

Divide 10 by -\frac{5}{2} by multiplying 10 by the reciprocal of -\frac{5}{2}.

$$t^{2}-4t=\frac{10}{-\frac{5}{2}}$$

Divide 10 by -\frac{5}{2} by multiplying 10 by the reciprocal of -\frac{5}{2}.

$$t^{2}-4t=-4$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add -4 to 4.

$$t^{2}-4t+4=0$$

Factor t^{2}-4t+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-2\right)^{2}}=\sqrt{0}$$

Simplify.

$$t-2=0 t-2=0$$

Add 2 to both sides of the equation.

$$t=2 t=2$$

The equation is now solved. Solutions are the same.

$$t=2$$