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x=5

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a+b=-10 ab=25

To solve the equation, factor x^{2}-10x+25 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

-1,-25 -5,-5

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 25.

-1-25=-26 -5-5=-10

Calculate the sum for each pair.

a=-5 b=-5

The solution is the pair that gives sum -10.

\left(x-5\right)\left(x-5\right)

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

\left(x-5\right)^{2}

Rewrite as a binomial square.

x=5

To find equation solution, solve x-5=0.

a+b=-10 ab=1\times 25=25

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

-1,-25 -5,-5

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 25.

-1-25=-26 -5-5=-10

Calculate the sum for each pair.

a=-5 b=-5

The solution is the pair that gives sum -10.

\left(x^{2}-5x\right)+\left(-5x+25\right)

Rewrite x^{2}-10x+25 as \left(x^{2}-5x\right)+\left(-5x+25\right).

x\left(x-5\right)-5\left(x-5\right)

Factor out x in the first and -5 in the second group.

\left(x-5\right)\left(x-5\right)

Factor out common term x-5 by using distributive property.

\left(x-5\right)^{2}

Rewrite as a binomial square.

x=5

To find equation solution, solve x-5=0.

x^{2}-10x+25=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(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-10\right)±\sqrt{100-4\times 25}}{2}

Square -10.

x=\frac{-\left(-10\right)±\sqrt{100-100}}{2}

Multiply -4 times 25.

x=\frac{-\left(-10\right)±\sqrt{0}}{2}

Add 100 to -100.

x=-\frac{-10}{2}

Take the square root of 0.

x=\frac{10}{2}

The opposite of -10 is 10.

x=5

Divide 10 by 2.

x^{2}-10x+25=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.

\left(x-5\right)^{2}=0

Factor x^{2}-10x+25. 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-5\right)^{2}}=\sqrt{0}

Take the square root of both sides of the equation.

x-5=0 x-5=0

Simplify.

x=5 x=5

Add 5 to both sides of the equation.

x=5

The equation is now solved. Solutions are the same.

x ^ 2 -10x +25 = 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 = 10 rs = 25

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 = 5 - u s = 5 + u

Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>

(5 - u) (5 + u) = 25

To solve for unknown quantity u, substitute these in the product equation rs = 25

25 - u^2 = 25

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = 25-25 = 0

Simplify the expression by subtracting 25 on both sides

u^2 = 0 u = 0

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r = s = 5

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.