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4t^{2}-1681t+32400=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{-\left(-1681\right)±\sqrt{\left(-1681\right)^{2}-4\times 4\times 32400}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -1681 for b, and 32400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-1681\right)±\sqrt{2825761-4\times 4\times 32400}}{2\times 4}
Square -1681.
t=\frac{-\left(-1681\right)±\sqrt{2825761-16\times 32400}}{2\times 4}
Multiply -4 times 4.
t=\frac{-\left(-1681\right)±\sqrt{2825761-518400}}{2\times 4}
Multiply -16 times 32400.
t=\frac{-\left(-1681\right)±\sqrt{2307361}}{2\times 4}
Add 2825761 to -518400.
t=\frac{-\left(-1681\right)±1519}{2\times 4}
Take the square root of 2307361.
t=\frac{1681±1519}{2\times 4}
The opposite of -1681 is 1681.
t=\frac{1681±1519}{8}
Multiply 2 times 4.
t=\frac{3200}{8}
Now solve the equation t=\frac{1681±1519}{8} when ± is plus. Add 1681 to 1519.
t=400
Divide 3200 by 8.
t=\frac{162}{8}
Now solve the equation t=\frac{1681±1519}{8} when ± is minus. Subtract 1519 from 1681.
t=\frac{81}{4}
Reduce the fraction \frac{162}{8} to lowest terms by extracting and canceling out 2.
t=400 t=\frac{81}{4}
The equation is now solved.
4t^{2}-1681t+32400=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.
4t^{2}-1681t+32400-32400=-32400
Subtract 32400 from both sides of the equation.
4t^{2}-1681t=-32400
Subtracting 32400 from itself leaves 0.
\frac{4t^{2}-1681t}{4}=-\frac{32400}{4}
Divide both sides by 4.
t^{2}-\frac{1681}{4}t=-\frac{32400}{4}
Dividing by 4 undoes the multiplication by 4.
t^{2}-\frac{1681}{4}t=-8100
Divide -32400 by 4.
t^{2}-\frac{1681}{4}t+\left(-\frac{1681}{8}\right)^{2}=-8100+\left(-\frac{1681}{8}\right)^{2}
Divide -\frac{1681}{4}, the coefficient of the x term, by 2 to get -\frac{1681}{8}. Then add the square of -\frac{1681}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-\frac{1681}{4}t+\frac{2825761}{64}=-8100+\frac{2825761}{64}
Square -\frac{1681}{8} by squaring both the numerator and the denominator of the fraction.
t^{2}-\frac{1681}{4}t+\frac{2825761}{64}=\frac{2307361}{64}
Add -8100 to \frac{2825761}{64}.
\left(t-\frac{1681}{8}\right)^{2}=\frac{2307361}{64}
Factor t^{2}-\frac{1681}{4}t+\frac{2825761}{64}. 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{1681}{8}\right)^{2}}=\sqrt{\frac{2307361}{64}}
Take the square root of both sides of the equation.
t-\frac{1681}{8}=\frac{1519}{8} t-\frac{1681}{8}=-\frac{1519}{8}
Simplify.
t=400 t=\frac{81}{4}
Add \frac{1681}{8} to both sides of the equation.
x ^ 2 -\frac{1681}{4}x +8100 = 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.This is achieved by dividing both sides of the equation by 4
r + s = \frac{1681}{4} rs = 8100
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{1681}{8} - u s = \frac{1681}{8} + u
Two numbers r and s sum up to \frac{1681}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{1681}{4} = \frac{1681}{8}. 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{1681}{8} - u) (\frac{1681}{8} + u) = 8100
To solve for unknown quantity u, substitute these in the product equation rs = 8100
\frac{2825761}{64} - u^2 = 8100
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 8100-\frac{2825761}{64} = -\frac{2307361}{64}
Simplify the expression by subtracting \frac{2825761}{64} on both sides
u^2 = \frac{2307361}{64} u = \pm\sqrt{\frac{2307361}{64}} = \pm \frac{1519}{8}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1681}{8} - \frac{1519}{8} = 20.250 s = \frac{1681}{8} + \frac{1519}{8} = 400
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.