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-16t^{2}+80t+5=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
t=\frac{-80±\sqrt{80^{2}-4\left(-16\right)\times 5}}{2\left(-16\right)}
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{6400-4\left(-16\right)\times 5}}{2\left(-16\right)}
Square 80.
t=\frac{-80±\sqrt{6400+64\times 5}}{2\left(-16\right)}
Multiply -4 times -16.
t=\frac{-80±\sqrt{6400+320}}{2\left(-16\right)}
Multiply 64 times 5.
t=\frac{-80±\sqrt{6720}}{2\left(-16\right)}
Add 6400 to 320.
t=\frac{-80±8\sqrt{105}}{2\left(-16\right)}
Take the square root of 6720.
t=\frac{-80±8\sqrt{105}}{-32}
Multiply 2 times -16.
t=\frac{8\sqrt{105}-80}{-32}
Now solve the equation t=\frac{-80±8\sqrt{105}}{-32} when ± is plus. Add -80 to 8\sqrt{105}.
t=-\frac{\sqrt{105}}{4}+\frac{5}{2}
Divide -80+8\sqrt{105} by -32.
t=\frac{-8\sqrt{105}-80}{-32}
Now solve the equation t=\frac{-80±8\sqrt{105}}{-32} when ± is minus. Subtract 8\sqrt{105} from -80.
t=\frac{\sqrt{105}}{4}+\frac{5}{2}
Divide -80-8\sqrt{105} by -32.
-16t^{2}+80t+5=-16\left(t-\left(-\frac{\sqrt{105}}{4}+\frac{5}{2}\right)\right)\left(t-\left(\frac{\sqrt{105}}{4}+\frac{5}{2}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2}-\frac{\sqrt{105}}{4} for x_{1} and \frac{5}{2}+\frac{\sqrt{105}}{4} for x_{2}.
x ^ 2 -5x -\frac{5}{16} = 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 = -\frac{5}{16}
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) = -\frac{5}{16}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{16}
\frac{25}{4} - u^2 = -\frac{5}{16}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{5}{16}-\frac{25}{4} = -\frac{105}{16}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = \frac{105}{16} u = \pm\sqrt{\frac{105}{16}} = \pm \frac{\sqrt{105}}{4}
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{105}}{4} = -0.062 s = \frac{5}{2} + \frac{\sqrt{105}}{4} = 5.062
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.