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-49t^{2}+98t+100=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{-98±\sqrt{98^{2}-4\left(-49\right)\times 100}}{2\left(-49\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -49 for a, 98 for b, and 100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-98±\sqrt{9604-4\left(-49\right)\times 100}}{2\left(-49\right)}
Square 98.
t=\frac{-98±\sqrt{9604+196\times 100}}{2\left(-49\right)}
Multiply -4 times -49.
t=\frac{-98±\sqrt{9604+19600}}{2\left(-49\right)}
Multiply 196 times 100.
t=\frac{-98±\sqrt{29204}}{2\left(-49\right)}
Add 9604 to 19600.
t=\frac{-98±14\sqrt{149}}{2\left(-49\right)}
Take the square root of 29204.
t=\frac{-98±14\sqrt{149}}{-98}
Multiply 2 times -49.
t=\frac{14\sqrt{149}-98}{-98}
Now solve the equation t=\frac{-98±14\sqrt{149}}{-98} when ± is plus. Add -98 to 14\sqrt{149}.
t=-\frac{\sqrt{149}}{7}+1
Divide -98+14\sqrt{149} by -98.
t=\frac{-14\sqrt{149}-98}{-98}
Now solve the equation t=\frac{-98±14\sqrt{149}}{-98} when ± is minus. Subtract 14\sqrt{149} from -98.
t=\frac{\sqrt{149}}{7}+1
Divide -98-14\sqrt{149} by -98.
t=-\frac{\sqrt{149}}{7}+1 t=\frac{\sqrt{149}}{7}+1
The equation is now solved.
-49t^{2}+98t+100=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}+98t+100-100=-100
Subtract 100 from both sides of the equation.
-49t^{2}+98t=-100
Subtracting 100 from itself leaves 0.
\frac{-49t^{2}+98t}{-49}=-\frac{100}{-49}
Divide both sides by -49.
t^{2}+\frac{98}{-49}t=-\frac{100}{-49}
Dividing by -49 undoes the multiplication by -49.
t^{2}-2t=-\frac{100}{-49}
Divide 98 by -49.
t^{2}-2t=\frac{100}{49}
Divide -100 by -49.
t^{2}-2t+1=\frac{100}{49}+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=\frac{149}{49}
Add \frac{100}{49} to 1.
\left(t-1\right)^{2}=\frac{149}{49}
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{\frac{149}{49}}
Take the square root of both sides of the equation.
t-1=\frac{\sqrt{149}}{7} t-1=-\frac{\sqrt{149}}{7}
Simplify.
t=\frac{\sqrt{149}}{7}+1 t=-\frac{\sqrt{149}}{7}+1
Add 1 to both sides of the equation.
x ^ 2 -2x -\frac{100}{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 = 2 rs = -\frac{100}{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 = 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) = -\frac{100}{49}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{100}{49}
1 - u^2 = -\frac{100}{49}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{100}{49}-1 = -\frac{149}{49}
Simplify the expression by subtracting 1 on both sides
u^2 = \frac{149}{49} u = \pm\sqrt{\frac{149}{49}} = \pm \frac{\sqrt{149}}{7}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - \frac{\sqrt{149}}{7} = -0.744 s = 1 + \frac{\sqrt{149}}{7} = 2.744
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.