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y^{2}+8y+41=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.
y=\frac{-8±\sqrt{8^{2}-4\times 41}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and 41 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-8±\sqrt{64-4\times 41}}{2}
Square 8.
y=\frac{-8±\sqrt{64-164}}{2}
Multiply -4 times 41.
y=\frac{-8±\sqrt{-100}}{2}
Add 64 to -164.
y=\frac{-8±10i}{2}
Take the square root of -100.
y=\frac{-8+10i}{2}
Now solve the equation y=\frac{-8±10i}{2} when ± is plus. Add -8 to 10i.
y=-4+5i
Divide -8+10i by 2.
y=\frac{-8-10i}{2}
Now solve the equation y=\frac{-8±10i}{2} when ± is minus. Subtract 10i from -8.
y=-4-5i
Divide -8-10i by 2.
y=-4+5i y=-4-5i
The equation is now solved.
y^{2}+8y+41=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.
y^{2}+8y+41-41=-41
Subtract 41 from both sides of the equation.
y^{2}+8y=-41
Subtracting 41 from itself leaves 0.
y^{2}+8y+4^{2}=-41+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}+8y+16=-41+16
Square 4.
y^{2}+8y+16=-25
Add -41 to 16.
\left(y+4\right)^{2}=-25
Factor y^{2}+8y+16. 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(y+4\right)^{2}}=\sqrt{-25}
Take the square root of both sides of the equation.
y+4=5i y+4=-5i
Simplify.
y=-4+5i y=-4-5i
Subtract 4 from both sides of the equation.