Solve -5.2=10t-6t^2

$$-5.2 = 10t-6 { t }^{ 2 }$$

Solve for t

$$t=\frac{\sqrt{1405}}{30}+\frac{5}{6}\approx 2.082777654$$

$$t=-\frac{\sqrt{1405}}{30}+\frac{5}{6}\approx -0.416110988$$

Steps Using the Quadratic Formula

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

$$10t-6t^{2}=-5.2$$

Add 5.2 to both sides.

$$10t-6t^{2}+5.2=0$$

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.

$$-6t^{2}+10t+5.2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 10 for b, and 5.2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-10±\sqrt{10^{2}-4\left(-6\right)\times 5.2}}{2\left(-6\right)}$$

Square 10.

$$t=\frac{-10±\sqrt{100-4\left(-6\right)\times 5.2}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$t=\frac{-10±\sqrt{100+24\times 5.2}}{2\left(-6\right)}$$

Multiply 24 times 5.2.

$$t=\frac{-10±\sqrt{100+124.8}}{2\left(-6\right)}$$

Add 100 to 124.8.

$$t=\frac{-10±\sqrt{224.8}}{2\left(-6\right)}$$

Take the square root of 224.8.

$$t=\frac{-10±\frac{2\sqrt{1405}}{5}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$t=\frac{-10±\frac{2\sqrt{1405}}{5}}{-12}$$

Now solve the equation t=\frac{-10±\frac{2\sqrt{1405}}{5}}{-12} when ± is plus. Add -10 to \frac{2\sqrt{1405}}{5}.

$$t=\frac{\frac{2\sqrt{1405}}{5}-10}{-12}$$

Divide -10+\frac{2\sqrt{1405}}{5} by -12.

$$t=-\frac{\sqrt{1405}}{30}+\frac{5}{6}$$

Now solve the equation t=\frac{-10±\frac{2\sqrt{1405}}{5}}{-12} when ± is minus. Subtract \frac{2\sqrt{1405}}{5} from -10.

$$t=\frac{-\frac{2\sqrt{1405}}{5}-10}{-12}$$

Divide -10-\frac{2\sqrt{1405}}{5} by -12.

$$t=\frac{\sqrt{1405}}{30}+\frac{5}{6}$$

The equation is now solved.

$$t=-\frac{\sqrt{1405}}{30}+\frac{5}{6} t=\frac{\sqrt{1405}}{30}+\frac{5}{6}$$

Steps for Completing the Square

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

$$10t-6t^{2}=-5.2$$

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.

$$-6t^{2}+10t=-5.2$$

Divide both sides by -6.

$$\frac{-6t^{2}+10t}{-6}=-\frac{5.2}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$t^{2}+\frac{10}{-6}t=-\frac{5.2}{-6}$$

Reduce the fraction \frac{10}{-6} to lowest terms by extracting and canceling out 2.

$$t^{2}-\frac{5}{3}t=-\frac{5.2}{-6}$$

Divide -5.2 by -6.

$$t^{2}-\frac{5}{3}t=\frac{13}{15}$$

Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{5}{3}t+\left(-\frac{5}{6}\right)^{2}=\frac{13}{15}+\left(-\frac{5}{6}\right)^{2}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{5}{3}t+\frac{25}{36}=\frac{13}{15}+\frac{25}{36}$$

Add \frac{13}{15} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{5}{3}t+\frac{25}{36}=\frac{281}{180}$$

Factor t^{2}-\frac{5}{3}t+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{5}{6}\right)^{2}=\frac{281}{180}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{5}{6}\right)^{2}}=\sqrt{\frac{281}{180}}$$

Simplify.

$$t-\frac{5}{6}=\frac{\sqrt{1405}}{30} t-\frac{5}{6}=-\frac{\sqrt{1405}}{30}$$

Add \frac{5}{6} to both sides of the equation.

$$t=\frac{\sqrt{1405}}{30}+\frac{5}{6} t=-\frac{\sqrt{1405}}{30}+\frac{5}{6}$$