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u^{2}-80u+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.
u=\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}.
u=\frac{-\left(-80\right)±\sqrt{6400-4\times 1200}}{2}
Square -80.
u=\frac{-\left(-80\right)±\sqrt{6400-4800}}{2}
Multiply -4 times 1200.
u=\frac{-\left(-80\right)±\sqrt{1600}}{2}
Add 6400 to -4800.
u=\frac{-\left(-80\right)±40}{2}
Take the square root of 1600.
u=\frac{80±40}{2}
The opposite of -80 is 80.
u=\frac{120}{2}
Now solve the equation u=\frac{80±40}{2} when ± is plus. Add 80 to 40.
u=60
Divide 120 by 2.
u=\frac{40}{2}
Now solve the equation u=\frac{80±40}{2} when ± is minus. Subtract 40 from 80.
u=20
Divide 40 by 2.
u=60 u=20
The equation is now solved.
u^{2}-80u+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.
u^{2}-80u+1200-1200=-1200
Subtract 1200 from both sides of the equation.
u^{2}-80u=-1200
Subtracting 1200 from itself leaves 0.
u^{2}-80u+\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.
u^{2}-80u+1600=-1200+1600
Square -40.
u^{2}-80u+1600=400
Add -1200 to 1600.
\left(u-40\right)^{2}=400
Factor u^{2}-80u+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(u-40\right)^{2}}=\sqrt{400}
Take the square root of both sides of the equation.
u-40=20 u-40=-20
Simplify.
u=60 u=20
Add 40 to both sides of the equation.
x ^ 2 -80x +1200 = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.
r + s = 80 rs = 1200
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = 40 - u s = 40 + u
Two numbers r and s sum up to 80 exactly when the average of the two numbers is \frac{1}{2}*80 = 40. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(40 - u) (40 + u) = 1200
To solve for unknown quantity u, substitute these in the product equation rs = 1200
1600 - u^2 = 1200
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1200-1600 = -400
Simplify the expression by subtracting 1600 on both sides
u^2 = 400 u = \pm\sqrt{400} = \pm 20
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =40 - 20 = 20 s = 40 + 20 = 60
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.