Solve for t
t=\sqrt{2}+2\approx 3.414213562
t=2-\sqrt{2}\approx 0.585786438
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-\frac{1}{2}t^{2}+2t=1
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{1}{2}t^{2}+2t-1=1-1
Subtract 1 from both sides of the equation.
-\frac{1}{2}t^{2}+2t-1=0
Subtracting 1 from itself leaves 0.
t=\frac{-2±\sqrt{2^{2}-4\left(-\frac{1}{2}\right)\left(-1\right)}}{2\left(-\frac{1}{2}\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{2} for a, 2 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-2±\sqrt{4-4\left(-\frac{1}{2}\right)\left(-1\right)}}{2\left(-\frac{1}{2}\right)}
Square 2.
t=\frac{-2±\sqrt{4+2\left(-1\right)}}{2\left(-\frac{1}{2}\right)}
Multiply -4 times -\frac{1}{2}.
t=\frac{-2±\sqrt{4-2}}{2\left(-\frac{1}{2}\right)}
Multiply 2 times -1.
t=\frac{-2±\sqrt{2}}{2\left(-\frac{1}{2}\right)}
Add 4 to -2.
t=\frac{-2±\sqrt{2}}{-1}
Multiply 2 times -\frac{1}{2}.
t=\frac{\sqrt{2}-2}{-1}
Now solve the equation t=\frac{-2±\sqrt{2}}{-1} when ± is plus. Add -2 to \sqrt{2}.
t=2-\sqrt{2}
Divide -2+\sqrt{2} by -1.
t=\frac{-\sqrt{2}-2}{-1}
Now solve the equation t=\frac{-2±\sqrt{2}}{-1} when ± is minus. Subtract \sqrt{2} from -2.
t=\sqrt{2}+2
Divide -2-\sqrt{2} by -1.
t=2-\sqrt{2} t=\sqrt{2}+2
The equation is now solved.
-\frac{1}{2}t^{2}+2t=1
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{-\frac{1}{2}t^{2}+2t}{-\frac{1}{2}}=\frac{1}{-\frac{1}{2}}
Multiply both sides by -2.
t^{2}+\frac{2}{-\frac{1}{2}}t=\frac{1}{-\frac{1}{2}}
Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.
t^{2}-4t=\frac{1}{-\frac{1}{2}}
Divide 2 by -\frac{1}{2} by multiplying 2 by the reciprocal of -\frac{1}{2}.
t^{2}-4t=-2
Divide 1 by -\frac{1}{2} by multiplying 1 by the reciprocal of -\frac{1}{2}.
t^{2}-4t+\left(-2\right)^{2}=-2+\left(-2\right)^{2}
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+4=-2+4
Square -2.
t^{2}-4t+4=2
Add -2 to 4.
\left(t-2\right)^{2}=2
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}.
\sqrt{\left(t-2\right)^{2}}=\sqrt{2}
Take the square root of both sides of the equation.
t-2=\sqrt{2} t-2=-\sqrt{2}
Simplify.
t=\sqrt{2}+2 t=2-\sqrt{2}
Add 2 to both sides of the equation.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}