Solve 5.25=10t+1/2*2t^2

$$5.25 = 10 t + \frac { 1 } { 2 } \times 2 t ^ { 2 }$$

Solve for t

$$t=-10.5$$

$$t=0.5$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 2 to get 1.

$$5.25=10t+t^{2}$$

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

$$10t+t^{2}=5.25$$

Subtract 5.25 from both sides.

$$10t+t^{2}-5.25=0$$

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.

$$t^{2}+10t-5.25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and -5.25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-10±\sqrt{10^{2}-4\left(-5.25\right)}}{2}$$

Square 10.

$$t=\frac{-10±\sqrt{100-4\left(-5.25\right)}}{2}$$

Multiply -4 times -5.25.

$$t=\frac{-10±\sqrt{100+21}}{2}$$

Add 100 to 21.

$$t=\frac{-10±\sqrt{121}}{2}$$

Take the square root of 121.

$$t=\frac{-10±11}{2}$$

Now solve the equation t=\frac{-10±11}{2} when ± is plus. Add -10 to 11.

$$t=\frac{1}{2}$$

Now solve the equation t=\frac{-10±11}{2} when ± is minus. Subtract 11 from -10.

$$t=-\frac{21}{2}$$

The equation is now solved.

$$t=\frac{1}{2} t=-\frac{21}{2}$$

Steps for Completing the Square

Multiply \frac{1}{2} and 2 to get 1.

$$5.25=10t+t^{2}$$

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

$$10t+t^{2}=5.25$$

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.

$$t^{2}+10t=5.25$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+10t+5^{2}=5.25+5^{2}$$

Square 5.

$$t^{2}+10t+25=5.25+25$$

Add 5.25 to 25.

$$t^{2}+10t+25=30.25$$

Factor t^{2}+10t+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+5\right)^{2}=30.25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+5\right)^{2}}=\sqrt{30.25}$$

Simplify.

$$t+5=\frac{11}{2} t+5=-\frac{11}{2}$$

Subtract 5 from both sides of the equation.

$$t=\frac{1}{2} t=-\frac{21}{2}$$