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5p^{2}-10p+10=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.
p=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 5\times 10}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -10 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
p=\frac{-\left(-10\right)±\sqrt{100-4\times 5\times 10}}{2\times 5}
Square -10.
p=\frac{-\left(-10\right)±\sqrt{100-20\times 10}}{2\times 5}
Multiply -4 times 5.
p=\frac{-\left(-10\right)±\sqrt{100-200}}{2\times 5}
Multiply -20 times 10.
p=\frac{-\left(-10\right)±\sqrt{-100}}{2\times 5}
Add 100 to -200.
p=\frac{-\left(-10\right)±10i}{2\times 5}
Take the square root of -100.
p=\frac{10±10i}{2\times 5}
The opposite of -10 is 10.
p=\frac{10±10i}{10}
Multiply 2 times 5.
p=\frac{10+10i}{10}
Now solve the equation p=\frac{10±10i}{10} when ± is plus. Add 10 to 10i.
p=1+i
Divide 10+10i by 10.
p=\frac{10-10i}{10}
Now solve the equation p=\frac{10±10i}{10} when ± is minus. Subtract 10i from 10.
p=1-i
Divide 10-10i by 10.
p=1+i p=1-i
The equation is now solved.
5p^{2}-10p+10=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.
5p^{2}-10p+10-10=-10
Subtract 10 from both sides of the equation.
5p^{2}-10p=-10
Subtracting 10 from itself leaves 0.
\frac{5p^{2}-10p}{5}=-\frac{10}{5}
Divide both sides by 5.
p^{2}+\left(-\frac{10}{5}\right)p=-\frac{10}{5}
Dividing by 5 undoes the multiplication by 5.
p^{2}-2p=-\frac{10}{5}
Divide -10 by 5.
p^{2}-2p=-2
Divide -10 by 5.
p^{2}-2p+1=-2+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
p^{2}-2p+1=-1
Add -2 to 1.
\left(p-1\right)^{2}=-1
Factor p^{2}-2p+1. 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(p-1\right)^{2}}=\sqrt{-1}
Take the square root of both sides of the equation.
p-1=i p-1=-i
Simplify.
p=1+i p=1-i
Add 1 to both sides of the equation.
x ^ 2 -2x +2 = 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.This is achieved by dividing both sides of the equation by 5
r + s = 2 rs = 2
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 = 1 - u s = 1 + u
Two numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{2}*2 = 1. 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>
(1 - u) (1 + u) = 2
To solve for unknown quantity u, substitute these in the product equation rs = 2
1 - u^2 = 2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 2-1 = 1
Simplify the expression by subtracting 1 on both sides
u^2 = -1 u = \pm\sqrt{-1} = \pm i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - i s = 1 + i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.