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x^{2}-80x+1200=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.
x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 1200}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -80 for b, and 1200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-80\right)±\sqrt{6400-4\times 1200}}{2}
Square -80.
x=\frac{-\left(-80\right)±\sqrt{6400-4800}}{2}
Multiply -4 times 1200.
x=\frac{-\left(-80\right)±\sqrt{1600}}{2}
Add 6400 to -4800.
x=\frac{-\left(-80\right)±40}{2}
Take the square root of 1600.
x=\frac{80±40}{2}
The opposite of -80 is 80.
x=\frac{120}{2}
Now solve the equation x=\frac{80±40}{2} when ± is plus. Add 80 to 40.
x=60
Divide 120 by 2.
x=\frac{40}{2}
Now solve the equation x=\frac{80±40}{2} when ± is minus. Subtract 40 from 80.
x=20
Divide 40 by 2.
x=60 x=20
The equation is now solved.
x^{2}-80x+1200=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.
x^{2}-80x+1200-1200=-1200
Subtract 1200 from both sides of the equation.
x^{2}-80x=-1200
Subtracting 1200 from itself leaves 0.
x^{2}-80x+\left(-40\right)^{2}=-1200+\left(-40\right)^{2}
Divide -80, the coefficient of the x term, by 2 to get -40. Then add the square of -40 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-80x+1600=-1200+1600
Square -40.
x^{2}-80x+1600=400
Add -1200 to 1600.
\left(x-40\right)^{2}=400
Factor x^{2}-80x+1600. 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(x-40\right)^{2}}=\sqrt{400}
Take the square root of both sides of the equation.
x-40=20 x-40=-20
Simplify.
x=60 x=20
Add 40 to both sides of the equation.