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2k^{2}-4k=5
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.
2k^{2}-4k-5=5-5
Subtract 5 from both sides of the equation.
2k^{2}-4k-5=0
Subtracting 5 from itself leaves 0.
k=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-5\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-5\right)}}{2\times 2}
Square -4.
k=\frac{-\left(-4\right)±\sqrt{16-8\left(-5\right)}}{2\times 2}
Multiply -4 times 2.
k=\frac{-\left(-4\right)±\sqrt{16+40}}{2\times 2}
Multiply -8 times -5.
k=\frac{-\left(-4\right)±\sqrt{56}}{2\times 2}
Add 16 to 40.
k=\frac{-\left(-4\right)±2\sqrt{14}}{2\times 2}
Take the square root of 56.
k=\frac{4±2\sqrt{14}}{2\times 2}
The opposite of -4 is 4.
k=\frac{4±2\sqrt{14}}{4}
Multiply 2 times 2.
k=\frac{2\sqrt{14}+4}{4}
Now solve the equation k=\frac{4±2\sqrt{14}}{4} when ± is plus. Add 4 to 2\sqrt{14}.
k=\frac{\sqrt{14}}{2}+1
Divide 4+2\sqrt{14} by 4.
k=\frac{4-2\sqrt{14}}{4}
Now solve the equation k=\frac{4±2\sqrt{14}}{4} when ± is minus. Subtract 2\sqrt{14} from 4.
k=-\frac{\sqrt{14}}{2}+1
Divide 4-2\sqrt{14} by 4.
k=\frac{\sqrt{14}}{2}+1 k=-\frac{\sqrt{14}}{2}+1
The equation is now solved.
2k^{2}-4k=5
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.
\frac{2k^{2}-4k}{2}=\frac{5}{2}
Divide both sides by 2.
k^{2}+\left(-\frac{4}{2}\right)k=\frac{5}{2}
Dividing by 2 undoes the multiplication by 2.
k^{2}-2k=\frac{5}{2}
Divide -4 by 2.
k^{2}-2k+1=\frac{5}{2}+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}-2k+1=\frac{7}{2}
Add \frac{5}{2} to 1.
\left(k-1\right)^{2}=\frac{7}{2}
Factor k^{2}-2k+1. 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-1\right)^{2}}=\sqrt{\frac{7}{2}}
Take the square root of both sides of the equation.
k-1=\frac{\sqrt{14}}{2} k-1=-\frac{\sqrt{14}}{2}
Simplify.
k=\frac{\sqrt{14}}{2}+1 k=-\frac{\sqrt{14}}{2}+1
Add 1 to both sides of the equation.