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-49t^{2}+2t-10=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(-49\right)\left(-10\right)}}{2\left(-49\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -49 for a, 2 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-2±\sqrt{4-4\left(-49\right)\left(-10\right)}}{2\left(-49\right)}
Square 2.
t=\frac{-2±\sqrt{4+196\left(-10\right)}}{2\left(-49\right)}
Multiply -4 times -49.
t=\frac{-2±\sqrt{4-1960}}{2\left(-49\right)}
Multiply 196 times -10.
t=\frac{-2±\sqrt{-1956}}{2\left(-49\right)}
Add 4 to -1960.
t=\frac{-2±2\sqrt{489}i}{2\left(-49\right)}
Take the square root of -1956.
t=\frac{-2±2\sqrt{489}i}{-98}
Multiply 2 times -49.
t=\frac{-2+2\sqrt{489}i}{-98}
Now solve the equation t=\frac{-2±2\sqrt{489}i}{-98} when ± is plus. Add -2 to 2i\sqrt{489}.
t=\frac{-\sqrt{489}i+1}{49}
Divide -2+2i\sqrt{489} by -98.
t=\frac{-2\sqrt{489}i-2}{-98}
Now solve the equation t=\frac{-2±2\sqrt{489}i}{-98} when ± is minus. Subtract 2i\sqrt{489} from -2.
t=\frac{1+\sqrt{489}i}{49}
Divide -2-2i\sqrt{489} by -98.
t=\frac{-\sqrt{489}i+1}{49} t=\frac{1+\sqrt{489}i}{49}
The equation is now solved.
-49t^{2}+2t-10=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.
-49t^{2}+2t-10-\left(-10\right)=-\left(-10\right)
Add 10 to both sides of the equation.
-49t^{2}+2t=-\left(-10\right)
Subtracting -10 from itself leaves 0.
-49t^{2}+2t=10
Subtract -10 from 0.
\frac{-49t^{2}+2t}{-49}=\frac{10}{-49}
Divide both sides by -49.
t^{2}+\frac{2}{-49}t=\frac{10}{-49}
Dividing by -49 undoes the multiplication by -49.
t^{2}-\frac{2}{49}t=\frac{10}{-49}
Divide 2 by -49.
t^{2}-\frac{2}{49}t=-\frac{10}{49}
Divide 10 by -49.
t^{2}-\frac{2}{49}t+\left(-\frac{1}{49}\right)^{2}=-\frac{10}{49}+\left(-\frac{1}{49}\right)^{2}
Divide -\frac{2}{49}, the coefficient of the x term, by 2 to get -\frac{1}{49}. Then add the square of -\frac{1}{49} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-\frac{2}{49}t+\frac{1}{2401}=-\frac{10}{49}+\frac{1}{2401}
Square -\frac{1}{49} by squaring both the numerator and the denominator of the fraction.
t^{2}-\frac{2}{49}t+\frac{1}{2401}=-\frac{489}{2401}
Add -\frac{10}{49} to \frac{1}{2401} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(t-\frac{1}{49}\right)^{2}=-\frac{489}{2401}
Factor t^{2}-\frac{2}{49}t+\frac{1}{2401}. 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-\frac{1}{49}\right)^{2}}=\sqrt{-\frac{489}{2401}}
Take the square root of both sides of the equation.
t-\frac{1}{49}=\frac{\sqrt{489}i}{49} t-\frac{1}{49}=-\frac{\sqrt{489}i}{49}
Simplify.
t=\frac{1+\sqrt{489}i}{49} t=\frac{-\sqrt{489}i+1}{49}
Add \frac{1}{49} to both sides of the equation.
x ^ 2 -\frac{2}{49}x +\frac{10}{49} = 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 = \frac{2}{49} rs = \frac{10}{49}
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 = \frac{1}{49} - u s = \frac{1}{49} + u
Two numbers r and s sum up to \frac{2}{49} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{49} = \frac{1}{49}. 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>
(\frac{1}{49} - u) (\frac{1}{49} + u) = \frac{10}{49}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{10}{49}
\frac{1}{2401} - u^2 = \frac{10}{49}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{10}{49}-\frac{1}{2401} = \frac{489}{2401}
Simplify the expression by subtracting \frac{1}{2401} on both sides
u^2 = -\frac{489}{2401} u = \pm\sqrt{-\frac{489}{2401}} = \pm \frac{\sqrt{489}}{49}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{49} - \frac{\sqrt{489}}{49}i = 0.020 - 0.451i s = \frac{1}{49} + \frac{\sqrt{489}}{49}i = 0.020 + 0.451i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.