Solve for x
x=10\sqrt{13}+40\approx 76.055512755
x=40-10\sqrt{13}\approx 3.944487245
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x^{2}-80x+300=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 300}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -80 for b, and 300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-80\right)±\sqrt{6400-4\times 300}}{2}
Square -80.
x=\frac{-\left(-80\right)±\sqrt{6400-1200}}{2}
Multiply -4 times 300.
x=\frac{-\left(-80\right)±\sqrt{5200}}{2}
Add 6400 to -1200.
x=\frac{-\left(-80\right)±20\sqrt{13}}{2}
Take the square root of 5200.
x=\frac{80±20\sqrt{13}}{2}
The opposite of -80 is 80.
x=\frac{20\sqrt{13}+80}{2}
Now solve the equation x=\frac{80±20\sqrt{13}}{2} when ± is plus. Add 80 to 20\sqrt{13}.
x=10\sqrt{13}+40
Divide 80+20\sqrt{13} by 2.
x=\frac{80-20\sqrt{13}}{2}
Now solve the equation x=\frac{80±20\sqrt{13}}{2} when ± is minus. Subtract 20\sqrt{13} from 80.
x=40-10\sqrt{13}
Divide 80-20\sqrt{13} by 2.
x=10\sqrt{13}+40 x=40-10\sqrt{13}
The equation is now solved.
x^{2}-80x+300=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+300-300=-300
Subtract 300 from both sides of the equation.
x^{2}-80x=-300
Subtracting 300 from itself leaves 0.
x^{2}-80x+\left(-40\right)^{2}=-300+\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=-300+1600
Square -40.
x^{2}-80x+1600=1300
Add -300 to 1600.
\left(x-40\right)^{2}=1300
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{1300}
Take the square root of both sides of the equation.
x-40=10\sqrt{13} x-40=-10\sqrt{13}
Simplify.
x=10\sqrt{13}+40 x=40-10\sqrt{13}
Add 40 to both sides of the equation.
x ^ 2 -80x +300 = 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 = 300
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) = 300
To solve for unknown quantity u, substitute these in the product equation rs = 300
1600 - u^2 = 300
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 300-1600 = -1300
Simplify the expression by subtracting 1600 on both sides
u^2 = 1300 u = \pm\sqrt{1300} = \pm \sqrt{1300}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =40 - \sqrt{1300} = 3.944 s = 40 + \sqrt{1300} = 76.056
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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