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f^{2}-3f-5=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.
f=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-5\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
f=\frac{-\left(-3\right)±\sqrt{9-4\left(-5\right)}}{2}
Square -3.
f=\frac{-\left(-3\right)±\sqrt{9+20}}{2}
Multiply -4 times -5.
f=\frac{-\left(-3\right)±\sqrt{29}}{2}
Add 9 to 20.
f=\frac{3±\sqrt{29}}{2}
The opposite of -3 is 3.
f=\frac{\sqrt{29}+3}{2}
Now solve the equation f=\frac{3±\sqrt{29}}{2} when ± is plus. Add 3 to \sqrt{29}.
f=\frac{3-\sqrt{29}}{2}
Now solve the equation f=\frac{3±\sqrt{29}}{2} when ± is minus. Subtract \sqrt{29} from 3.
f=\frac{\sqrt{29}+3}{2} f=\frac{3-\sqrt{29}}{2}
The equation is now solved.
f^{2}-3f-5=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.
f^{2}-3f-5-\left(-5\right)=-\left(-5\right)
Add 5 to both sides of the equation.
f^{2}-3f=-\left(-5\right)
Subtracting -5 from itself leaves 0.
f^{2}-3f=5
Subtract -5 from 0.
f^{2}-3f+\left(-\frac{3}{2}\right)^{2}=5+\left(-\frac{3}{2}\right)^{2}
Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
f^{2}-3f+\frac{9}{4}=5+\frac{9}{4}
Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.
f^{2}-3f+\frac{9}{4}=\frac{29}{4}
Add 5 to \frac{9}{4}.
\left(f-\frac{3}{2}\right)^{2}=\frac{29}{4}
Factor f^{2}-3f+\frac{9}{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(f-\frac{3}{2}\right)^{2}}=\sqrt{\frac{29}{4}}
Take the square root of both sides of the equation.
f-\frac{3}{2}=\frac{\sqrt{29}}{2} f-\frac{3}{2}=-\frac{\sqrt{29}}{2}
Simplify.
f=\frac{\sqrt{29}+3}{2} f=\frac{3-\sqrt{29}}{2}
Add \frac{3}{2} to both sides of the equation.
x ^ 2 -3x -5 = 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 = 3 rs = -5
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{3}{2} - u s = \frac{3}{2} + u
Two numbers r and s sum up to 3 exactly when the average of the two numbers is \frac{1}{2}*3 = \frac{3}{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.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{3}{2} - u) (\frac{3}{2} + u) = -5
To solve for unknown quantity u, substitute these in the product equation rs = -5
\frac{9}{4} - u^2 = -5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -5-\frac{9}{4} = -\frac{29}{4}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^2 = \frac{29}{4} u = \pm\sqrt{\frac{29}{4}} = \pm \frac{\sqrt{29}}{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{3}{2} - \frac{\sqrt{29}}{2} = -1.193 s = \frac{3}{2} + \frac{\sqrt{29}}{2} = 4.193
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.