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Solve for k (complex solution)
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Solve for k
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k^{2}+6k-80=-3
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}+6k-80-\left(-3\right)=-3-\left(-3\right)
Add 3 to both sides of the equation.
k^{2}+6k-80-\left(-3\right)=0
Subtracting -3 from itself leaves 0.
k^{2}+6k-77=0
Subtract -3 from -80.
k=\frac{-6±\sqrt{6^{2}-4\left(-77\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -77 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-6±\sqrt{36-4\left(-77\right)}}{2}
Square 6.
k=\frac{-6±\sqrt{36+308}}{2}
Multiply -4 times -77.
k=\frac{-6±\sqrt{344}}{2}
Add 36 to 308.
k=\frac{-6±2\sqrt{86}}{2}
Take the square root of 344.
k=\frac{2\sqrt{86}-6}{2}
Now solve the equation k=\frac{-6±2\sqrt{86}}{2} when ± is plus. Add -6 to 2\sqrt{86}.
k=\sqrt{86}-3
Divide -6+2\sqrt{86} by 2.
k=\frac{-2\sqrt{86}-6}{2}
Now solve the equation k=\frac{-6±2\sqrt{86}}{2} when ± is minus. Subtract 2\sqrt{86} from -6.
k=-\sqrt{86}-3
Divide -6-2\sqrt{86} by 2.
k=\sqrt{86}-3 k=-\sqrt{86}-3
The equation is now solved.
k^{2}+6k-80=-3
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}+6k-80-\left(-80\right)=-3-\left(-80\right)
Add 80 to both sides of the equation.
k^{2}+6k=-3-\left(-80\right)
Subtracting -80 from itself leaves 0.
k^{2}+6k=77
Subtract -80 from -3.
k^{2}+6k+3^{2}=77+3^{2}
Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}+6k+9=77+9
Square 3.
k^{2}+6k+9=86
Add 77 to 9.
\left(k+3\right)^{2}=86
Factor k^{2}+6k+9. 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+3\right)^{2}}=\sqrt{86}
Take the square root of both sides of the equation.
k+3=\sqrt{86} k+3=-\sqrt{86}
Simplify.
k=\sqrt{86}-3 k=-\sqrt{86}-3
Subtract 3 from both sides of the equation.
k^{2}+6k-80=-3
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}+6k-80-\left(-3\right)=-3-\left(-3\right)
Add 3 to both sides of the equation.
k^{2}+6k-80-\left(-3\right)=0
Subtracting -3 from itself leaves 0.
k^{2}+6k-77=0
Subtract -3 from -80.
k=\frac{-6±\sqrt{6^{2}-4\left(-77\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -77 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-6±\sqrt{36-4\left(-77\right)}}{2}
Square 6.
k=\frac{-6±\sqrt{36+308}}{2}
Multiply -4 times -77.
k=\frac{-6±\sqrt{344}}{2}
Add 36 to 308.
k=\frac{-6±2\sqrt{86}}{2}
Take the square root of 344.
k=\frac{2\sqrt{86}-6}{2}
Now solve the equation k=\frac{-6±2\sqrt{86}}{2} when ± is plus. Add -6 to 2\sqrt{86}.
k=\sqrt{86}-3
Divide -6+2\sqrt{86} by 2.
k=\frac{-2\sqrt{86}-6}{2}
Now solve the equation k=\frac{-6±2\sqrt{86}}{2} when ± is minus. Subtract 2\sqrt{86} from -6.
k=-\sqrt{86}-3
Divide -6-2\sqrt{86} by 2.
k=\sqrt{86}-3 k=-\sqrt{86}-3
The equation is now solved.
k^{2}+6k-80=-3
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}+6k-80-\left(-80\right)=-3-\left(-80\right)
Add 80 to both sides of the equation.
k^{2}+6k=-3-\left(-80\right)
Subtracting -80 from itself leaves 0.
k^{2}+6k=77
Subtract -80 from -3.
k^{2}+6k+3^{2}=77+3^{2}
Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}+6k+9=77+9
Square 3.
k^{2}+6k+9=86
Add 77 to 9.
\left(k+3\right)^{2}=86
Factor k^{2}+6k+9. 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+3\right)^{2}}=\sqrt{86}
Take the square root of both sides of the equation.
k+3=\sqrt{86} k+3=-\sqrt{86}
Simplify.
k=\sqrt{86}-3 k=-\sqrt{86}-3
Subtract 3 from both sides of the equation.