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-t^{2}+2t+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.
t=\frac{-2±\sqrt{2^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 2 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-2±\sqrt{4-4\left(-1\right)\times 2}}{2\left(-1\right)}
Square 2.
t=\frac{-2±\sqrt{4+4\times 2}}{2\left(-1\right)}
Multiply -4 times -1.
t=\frac{-2±\sqrt{4+8}}{2\left(-1\right)}
Multiply 4 times 2.
t=\frac{-2±\sqrt{12}}{2\left(-1\right)}
Add 4 to 8.
t=\frac{-2±2\sqrt{3}}{2\left(-1\right)}
Take the square root of 12.
t=\frac{-2±2\sqrt{3}}{-2}
Multiply 2 times -1.
t=\frac{2\sqrt{3}-2}{-2}
Now solve the equation t=\frac{-2±2\sqrt{3}}{-2} when ± is plus. Add -2 to 2\sqrt{3}.
t=1-\sqrt{3}
Divide -2+2\sqrt{3} by -2.
t=\frac{-2\sqrt{3}-2}{-2}
Now solve the equation t=\frac{-2±2\sqrt{3}}{-2} when ± is minus. Subtract 2\sqrt{3} from -2.
t=\sqrt{3}+1
Divide -2-2\sqrt{3} by -2.
t=1-\sqrt{3} t=\sqrt{3}+1
The equation is now solved.
-t^{2}+2t+2=0
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}+2t+2-2=-2
Subtract 2 from both sides of the equation.
-t^{2}+2t=-2
Subtracting 2 from itself leaves 0.
\frac{-t^{2}+2t}{-1}=-\frac{2}{-1}
Divide both sides by -1.
t^{2}+\frac{2}{-1}t=-\frac{2}{-1}
Dividing by -1 undoes the multiplication by -1.
t^{2}-2t=-\frac{2}{-1}
Divide 2 by -1.
t^{2}-2t=2
Divide -2 by -1.
t^{2}-2t+1=2+1
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=3
Add 2 to 1.
\left(t-1\right)^{2}=3
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}.
\sqrt{\left(t-1\right)^{2}}=\sqrt{3}
Take the square root of both sides of the equation.
t-1=\sqrt{3} t-1=-\sqrt{3}
Simplify.
t=\sqrt{3}+1 t=1-\sqrt{3}
Add 1 to both sides of the equation.
x ^ 2 -2x -2 = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.
r + s = 2 rs = -2
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = 1 - u s = 1 + u
Two numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{2}*2 = 1. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(1 - u) (1 + u) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
1 - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-1 = -3
Simplify the expression by subtracting 1 on both sides
u^2 = 3 u = \pm\sqrt{3} = \pm \sqrt{3}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - \sqrt{3} = -0.732 s = 1 + \sqrt{3} = 2.732
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.