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5k^{2}-16k-54=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.
k=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\left(-54\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -16 for b, and -54 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-\left(-16\right)±\sqrt{256-4\times 5\left(-54\right)}}{2\times 5}
Square -16.
k=\frac{-\left(-16\right)±\sqrt{256-20\left(-54\right)}}{2\times 5}
Multiply -4 times 5.
k=\frac{-\left(-16\right)±\sqrt{256+1080}}{2\times 5}
Multiply -20 times -54.
k=\frac{-\left(-16\right)±\sqrt{1336}}{2\times 5}
Add 256 to 1080.
k=\frac{-\left(-16\right)±2\sqrt{334}}{2\times 5}
Take the square root of 1336.
k=\frac{16±2\sqrt{334}}{2\times 5}
The opposite of -16 is 16.
k=\frac{16±2\sqrt{334}}{10}
Multiply 2 times 5.
k=\frac{2\sqrt{334}+16}{10}
Now solve the equation k=\frac{16±2\sqrt{334}}{10} when ± is plus. Add 16 to 2\sqrt{334}.
k=\frac{\sqrt{334}+8}{5}
Divide 16+2\sqrt{334} by 10.
k=\frac{16-2\sqrt{334}}{10}
Now solve the equation k=\frac{16±2\sqrt{334}}{10} when ± is minus. Subtract 2\sqrt{334} from 16.
k=\frac{8-\sqrt{334}}{5}
Divide 16-2\sqrt{334} by 10.
k=\frac{\sqrt{334}+8}{5} k=\frac{8-\sqrt{334}}{5}
The equation is now solved.
5k^{2}-16k-54=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.
5k^{2}-16k-54-\left(-54\right)=-\left(-54\right)
Add 54 to both sides of the equation.
5k^{2}-16k=-\left(-54\right)
Subtracting -54 from itself leaves 0.
5k^{2}-16k=54
Subtract -54 from 0.
\frac{5k^{2}-16k}{5}=\frac{54}{5}
Divide both sides by 5.
k^{2}-\frac{16}{5}k=\frac{54}{5}
Dividing by 5 undoes the multiplication by 5.
k^{2}-\frac{16}{5}k+\left(-\frac{8}{5}\right)^{2}=\frac{54}{5}+\left(-\frac{8}{5}\right)^{2}
Divide -\frac{16}{5}, the coefficient of the x term, by 2 to get -\frac{8}{5}. Then add the square of -\frac{8}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}-\frac{16}{5}k+\frac{64}{25}=\frac{54}{5}+\frac{64}{25}
Square -\frac{8}{5} by squaring both the numerator and the denominator of the fraction.
k^{2}-\frac{16}{5}k+\frac{64}{25}=\frac{334}{25}
Add \frac{54}{5} to \frac{64}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(k-\frac{8}{5}\right)^{2}=\frac{334}{25}
Factor k^{2}-\frac{16}{5}k+\frac{64}{25}. 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(k-\frac{8}{5}\right)^{2}}=\sqrt{\frac{334}{25}}
Take the square root of both sides of the equation.
k-\frac{8}{5}=\frac{\sqrt{334}}{5} k-\frac{8}{5}=-\frac{\sqrt{334}}{5}
Simplify.
k=\frac{\sqrt{334}+8}{5} k=\frac{8-\sqrt{334}}{5}
Add \frac{8}{5} to both sides of the equation.
x ^ 2 -\frac{16}{5}x -\frac{54}{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.This is achieved by dividing both sides of the equation by 5
r + s = \frac{16}{5} rs = -\frac{54}{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{8}{5} - u s = \frac{8}{5} + u
Two numbers r and s sum up to \frac{16}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{16}{5} = \frac{8}{5}. 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{8}{5} - u) (\frac{8}{5} + u) = -\frac{54}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{54}{5}
\frac{64}{25} - u^2 = -\frac{54}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{54}{5}-\frac{64}{25} = -\frac{334}{25}
Simplify the expression by subtracting \frac{64}{25} on both sides
u^2 = \frac{334}{25} u = \pm\sqrt{\frac{334}{25}} = \pm \frac{\sqrt{334}}{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{8}{5} - \frac{\sqrt{334}}{5} = -2.055 s = \frac{8}{5} + \frac{\sqrt{334}}{5} = 5.255
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.