Solve for t
t = \frac{\sqrt{277} + 5}{14} \approx 1.545951213
t=\frac{5-\sqrt{277}}{14}\approx -0.831665498
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7t^{2}-5t-9=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 7\left(-9\right)}}{2\times 7}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -5 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-5\right)±\sqrt{25-4\times 7\left(-9\right)}}{2\times 7}
Square -5.
t=\frac{-\left(-5\right)±\sqrt{25-28\left(-9\right)}}{2\times 7}
Multiply -4 times 7.
t=\frac{-\left(-5\right)±\sqrt{25+252}}{2\times 7}
Multiply -28 times -9.
t=\frac{-\left(-5\right)±\sqrt{277}}{2\times 7}
Add 25 to 252.
t=\frac{5±\sqrt{277}}{2\times 7}
The opposite of -5 is 5.
t=\frac{5±\sqrt{277}}{14}
Multiply 2 times 7.
t=\frac{\sqrt{277}+5}{14}
Now solve the equation t=\frac{5±\sqrt{277}}{14} when ± is plus. Add 5 to \sqrt{277}.
t=\frac{5-\sqrt{277}}{14}
Now solve the equation t=\frac{5±\sqrt{277}}{14} when ± is minus. Subtract \sqrt{277} from 5.
t=\frac{\sqrt{277}+5}{14} t=\frac{5-\sqrt{277}}{14}
The equation is now solved.
7t^{2}-5t-9=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.
7t^{2}-5t-9-\left(-9\right)=-\left(-9\right)
Add 9 to both sides of the equation.
7t^{2}-5t=-\left(-9\right)
Subtracting -9 from itself leaves 0.
7t^{2}-5t=9
Subtract -9 from 0.
\frac{7t^{2}-5t}{7}=\frac{9}{7}
Divide both sides by 7.
t^{2}-\frac{5}{7}t=\frac{9}{7}
Dividing by 7 undoes the multiplication by 7.
t^{2}-\frac{5}{7}t+\left(-\frac{5}{14}\right)^{2}=\frac{9}{7}+\left(-\frac{5}{14}\right)^{2}
Divide -\frac{5}{7}, the coefficient of the x term, by 2 to get -\frac{5}{14}. Then add the square of -\frac{5}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-\frac{5}{7}t+\frac{25}{196}=\frac{9}{7}+\frac{25}{196}
Square -\frac{5}{14} by squaring both the numerator and the denominator of the fraction.
t^{2}-\frac{5}{7}t+\frac{25}{196}=\frac{277}{196}
Add \frac{9}{7} to \frac{25}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(t-\frac{5}{14}\right)^{2}=\frac{277}{196}
Factor t^{2}-\frac{5}{7}t+\frac{25}{196}. 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}{14}\right)^{2}}=\sqrt{\frac{277}{196}}
Take the square root of both sides of the equation.
t-\frac{5}{14}=\frac{\sqrt{277}}{14} t-\frac{5}{14}=-\frac{\sqrt{277}}{14}
Simplify.
t=\frac{\sqrt{277}+5}{14} t=\frac{5-\sqrt{277}}{14}
Add \frac{5}{14} to both sides of the equation.
x ^ 2 -\frac{5}{7}x -\frac{9}{7} = 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 7
r + s = \frac{5}{7} rs = -\frac{9}{7}
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}{14} - u s = \frac{5}{14} + u
Two numbers r and s sum up to \frac{5}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{7} = \frac{5}{14}. 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}{14} - u) (\frac{5}{14} + u) = -\frac{9}{7}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{7}
\frac{25}{196} - u^2 = -\frac{9}{7}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{9}{7}-\frac{25}{196} = -\frac{277}{196}
Simplify the expression by subtracting \frac{25}{196} on both sides
u^2 = \frac{277}{196} u = \pm\sqrt{\frac{277}{196}} = \pm \frac{\sqrt{277}}{14}
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}{14} - \frac{\sqrt{277}}{14} = -0.832 s = \frac{5}{14} + \frac{\sqrt{277}}{14} = 1.546
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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