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x^{2}-20x-8=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(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-8\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -20 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-20\right)±\sqrt{400-4\left(-8\right)}}{2}
Square -20.
x=\frac{-\left(-20\right)±\sqrt{400+32}}{2}
Multiply -4 times -8.
x=\frac{-\left(-20\right)±\sqrt{432}}{2}
Add 400 to 32.
x=\frac{-\left(-20\right)±12\sqrt{3}}{2}
Take the square root of 432.
x=\frac{20±12\sqrt{3}}{2}
The opposite of -20 is 20.
x=\frac{12\sqrt{3}+20}{2}
Now solve the equation x=\frac{20±12\sqrt{3}}{2} when ± is plus. Add 20 to 12\sqrt{3}.
x=6\sqrt{3}+10
Divide 20+12\sqrt{3} by 2.
x=\frac{20-12\sqrt{3}}{2}
Now solve the equation x=\frac{20±12\sqrt{3}}{2} when ± is minus. Subtract 12\sqrt{3} from 20.
x=10-6\sqrt{3}
Divide 20-12\sqrt{3} by 2.
x=6\sqrt{3}+10 x=10-6\sqrt{3}
The equation is now solved.
x^{2}-20x-8=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}-20x-8-\left(-8\right)=-\left(-8\right)
Add 8 to both sides of the equation.
x^{2}-20x=-\left(-8\right)
Subtracting -8 from itself leaves 0.
x^{2}-20x=8
Subtract -8 from 0.
x^{2}-20x+\left(-10\right)^{2}=8+\left(-10\right)^{2}
Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-20x+100=8+100
Square -10.
x^{2}-20x+100=108
Add 8 to 100.
\left(x-10\right)^{2}=108
Factor x^{2}-20x+100. 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-10\right)^{2}}=\sqrt{108}
Take the square root of both sides of the equation.
x-10=6\sqrt{3} x-10=-6\sqrt{3}
Simplify.
x=6\sqrt{3}+10 x=10-6\sqrt{3}
Add 10 to both sides of the equation.
x ^ 2 -20x -8 = 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 = 20 rs = -8
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 = 10 - u s = 10 + u
Two numbers r and s sum up to 20 exactly when the average of the two numbers is \frac{1}{2}*20 = 10. 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>
(10 - u) (10 + u) = -8
To solve for unknown quantity u, substitute these in the product equation rs = -8
100 - u^2 = -8
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -8-100 = -108
Simplify the expression by subtracting 100 on both sides
u^2 = 108 u = \pm\sqrt{108} = \pm \sqrt{108}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =10 - \sqrt{108} = -0.392 s = 10 + \sqrt{108} = 20.392
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.