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8t^{2}-5t-375=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 8\left(-375\right)}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -5 for b, and -375 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-5\right)±\sqrt{25-4\times 8\left(-375\right)}}{2\times 8}
Square -5.
t=\frac{-\left(-5\right)±\sqrt{25-32\left(-375\right)}}{2\times 8}
Multiply -4 times 8.
t=\frac{-\left(-5\right)±\sqrt{25+12000}}{2\times 8}
Multiply -32 times -375.
t=\frac{-\left(-5\right)±\sqrt{12025}}{2\times 8}
Add 25 to 12000.
t=\frac{-\left(-5\right)±5\sqrt{481}}{2\times 8}
Take the square root of 12025.
t=\frac{5±5\sqrt{481}}{2\times 8}
The opposite of -5 is 5.
t=\frac{5±5\sqrt{481}}{16}
Multiply 2 times 8.
t=\frac{5\sqrt{481}+5}{16}
Now solve the equation t=\frac{5±5\sqrt{481}}{16} when ± is plus. Add 5 to 5\sqrt{481}.
t=\frac{5-5\sqrt{481}}{16}
Now solve the equation t=\frac{5±5\sqrt{481}}{16} when ± is minus. Subtract 5\sqrt{481} from 5.
t=\frac{5\sqrt{481}+5}{16} t=\frac{5-5\sqrt{481}}{16}
The equation is now solved.
8t^{2}-5t-375=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.
8t^{2}-5t-375-\left(-375\right)=-\left(-375\right)
Add 375 to both sides of the equation.
8t^{2}-5t=-\left(-375\right)
Subtracting -375 from itself leaves 0.
8t^{2}-5t=375
Subtract -375 from 0.
\frac{8t^{2}-5t}{8}=\frac{375}{8}
Divide both sides by 8.
t^{2}-\frac{5}{8}t=\frac{375}{8}
Dividing by 8 undoes the multiplication by 8.
t^{2}-\frac{5}{8}t+\left(-\frac{5}{16}\right)^{2}=\frac{375}{8}+\left(-\frac{5}{16}\right)^{2}
Divide -\frac{5}{8}, the coefficient of the x term, by 2 to get -\frac{5}{16}. Then add the square of -\frac{5}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-\frac{5}{8}t+\frac{25}{256}=\frac{375}{8}+\frac{25}{256}
Square -\frac{5}{16} by squaring both the numerator and the denominator of the fraction.
t^{2}-\frac{5}{8}t+\frac{25}{256}=\frac{12025}{256}
Add \frac{375}{8} to \frac{25}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(t-\frac{5}{16}\right)^{2}=\frac{12025}{256}
Factor t^{2}-\frac{5}{8}t+\frac{25}{256}. 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}{16}\right)^{2}}=\sqrt{\frac{12025}{256}}
Take the square root of both sides of the equation.
t-\frac{5}{16}=\frac{5\sqrt{481}}{16} t-\frac{5}{16}=-\frac{5\sqrt{481}}{16}
Simplify.
t=\frac{5\sqrt{481}+5}{16} t=\frac{5-5\sqrt{481}}{16}
Add \frac{5}{16} to both sides of the equation.
x ^ 2 -\frac{5}{8}x -\frac{375}{8} = 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 8
r + s = \frac{5}{8} rs = -\frac{375}{8}
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}{16} - u s = \frac{5}{16} + u
Two numbers r and s sum up to \frac{5}{8} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{8} = \frac{5}{16}. 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{5}{16} - u) (\frac{5}{16} + u) = -\frac{375}{8}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{375}{8}
\frac{25}{256} - u^2 = -\frac{375}{8}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{375}{8}-\frac{25}{256} = -\frac{12025}{256}
Simplify the expression by subtracting \frac{25}{256} on both sides
u^2 = \frac{12025}{256} u = \pm\sqrt{\frac{12025}{256}} = \pm \frac{\sqrt{12025}}{16}
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}{16} - \frac{\sqrt{12025}}{16} = -6.541 s = \frac{5}{16} + \frac{\sqrt{12025}}{16} = 7.166
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.