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-16t^{2}+80t+64=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{-80±\sqrt{80^{2}-4\left(-16\right)\times 64}}{2\left(-16\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 80 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-80±\sqrt{6400-4\left(-16\right)\times 64}}{2\left(-16\right)}
Square 80.
t=\frac{-80±\sqrt{6400+64\times 64}}{2\left(-16\right)}
Multiply -4 times -16.
t=\frac{-80±\sqrt{6400+4096}}{2\left(-16\right)}
Multiply 64 times 64.
t=\frac{-80±\sqrt{10496}}{2\left(-16\right)}
Add 6400 to 4096.
t=\frac{-80±16\sqrt{41}}{2\left(-16\right)}
Take the square root of 10496.
t=\frac{-80±16\sqrt{41}}{-32}
Multiply 2 times -16.
t=\frac{16\sqrt{41}-80}{-32}
Now solve the equation t=\frac{-80±16\sqrt{41}}{-32} when ± is plus. Add -80 to 16\sqrt{41}.
t=\frac{5-\sqrt{41}}{2}
Divide -80+16\sqrt{41} by -32.
t=\frac{-16\sqrt{41}-80}{-32}
Now solve the equation t=\frac{-80±16\sqrt{41}}{-32} when ± is minus. Subtract 16\sqrt{41} from -80.
t=\frac{\sqrt{41}+5}{2}
Divide -80-16\sqrt{41} by -32.
t=\frac{5-\sqrt{41}}{2} t=\frac{\sqrt{41}+5}{2}
The equation is now solved.
-16t^{2}+80t+64=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.
-16t^{2}+80t+64-64=-64
Subtract 64 from both sides of the equation.
-16t^{2}+80t=-64
Subtracting 64 from itself leaves 0.
\frac{-16t^{2}+80t}{-16}=-\frac{64}{-16}
Divide both sides by -16.
t^{2}+\frac{80}{-16}t=-\frac{64}{-16}
Dividing by -16 undoes the multiplication by -16.
t^{2}-5t=-\frac{64}{-16}
Divide 80 by -16.
t^{2}-5t=4
Divide -64 by -16.
t^{2}-5t+\left(-\frac{5}{2}\right)^{2}=4+\left(-\frac{5}{2}\right)^{2}
Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-5t+\frac{25}{4}=4+\frac{25}{4}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
t^{2}-5t+\frac{25}{4}=\frac{41}{4}
Add 4 to \frac{25}{4}.
\left(t-\frac{5}{2}\right)^{2}=\frac{41}{4}
Factor t^{2}-5t+\frac{25}{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-\frac{5}{2}\right)^{2}}=\sqrt{\frac{41}{4}}
Take the square root of both sides of the equation.
t-\frac{5}{2}=\frac{\sqrt{41}}{2} t-\frac{5}{2}=-\frac{\sqrt{41}}{2}
Simplify.
t=\frac{\sqrt{41}+5}{2} t=\frac{5-\sqrt{41}}{2}
Add \frac{5}{2} to both sides of the equation.
x ^ 2 -5x -4 = 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 = 5 rs = -4
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{5}{2} - u s = \frac{5}{2} + u
Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{2}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{5}{2} - u) (\frac{5}{2} + u) = -4
To solve for unknown quantity u, substitute these in the product equation rs = -4
\frac{25}{4} - u^2 = -4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -4-\frac{25}{4} = -\frac{41}{4}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = \frac{41}{4} u = \pm\sqrt{\frac{41}{4}} = \pm \frac{\sqrt{41}}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{2} - \frac{\sqrt{41}}{2} = -0.702 s = \frac{5}{2} + \frac{\sqrt{41}}{2} = 5.702
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.