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k^{2}+10k=37
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^{2}+10k-37=37-37
Subtract 37 from both sides of the equation.
k^{2}+10k-37=0
Subtracting 37 from itself leaves 0.
k=\frac{-10±\sqrt{10^{2}-4\left(-37\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and -37 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-10±\sqrt{100-4\left(-37\right)}}{2}
Square 10.
k=\frac{-10±\sqrt{100+148}}{2}
Multiply -4 times -37.
k=\frac{-10±\sqrt{248}}{2}
Add 100 to 148.
k=\frac{-10±2\sqrt{62}}{2}
Take the square root of 248.
k=\frac{2\sqrt{62}-10}{2}
Now solve the equation k=\frac{-10±2\sqrt{62}}{2} when ± is plus. Add -10 to 2\sqrt{62}.
k=\sqrt{62}-5
Divide -10+2\sqrt{62} by 2.
k=\frac{-2\sqrt{62}-10}{2}
Now solve the equation k=\frac{-10±2\sqrt{62}}{2} when ± is minus. Subtract 2\sqrt{62} from -10.
k=-\sqrt{62}-5
Divide -10-2\sqrt{62} by 2.
k=\sqrt{62}-5 k=-\sqrt{62}-5
The equation is now solved.
k^{2}+10k=37
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.
k^{2}+10k+5^{2}=37+5^{2}
Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}+10k+25=37+25
Square 5.
k^{2}+10k+25=62
Add 37 to 25.
\left(k+5\right)^{2}=62
Factor k^{2}+10k+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+5\right)^{2}}=\sqrt{62}
Take the square root of both sides of the equation.
k+5=\sqrt{62} k+5=-\sqrt{62}
Simplify.
k=\sqrt{62}-5 k=-\sqrt{62}-5
Subtract 5 from both sides of the equation.