Solve for x
x=-5
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x^{2}+10x+25=0
Add 25 to both sides.
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 positive, a and b are both positive. 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.
x^{2}+10x+25=0
Add 25 to both sides.
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 positive, a and b are both positive. 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
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^{2}+10x-\left(-25\right)=-25-\left(-25\right)
Add 25 to both sides of the equation.
x^{2}+10x-\left(-25\right)=0
Subtracting -25 from itself leaves 0.
x^{2}+10x+25=0
Subtract -25 from 0.
x=\frac{-10±\sqrt{10^{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{-10±\sqrt{100-4\times 25}}{2}
Square 10.
x=\frac{-10±\sqrt{100-100}}{2}
Multiply -4 times 25.
x=\frac{-10±\sqrt{0}}{2}
Add 100 to -100.
x=-\frac{10}{2}
Take the square root of 0.
x=-5
Divide -10 by 2.
x^{2}+10x=-25
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}+10x+5^{2}=-25+5^{2}
Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+10x+25=-25+25
Square 5.
x^{2}+10x+25=0
Add -25 to 25.
\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
Subtract 5 from both sides of the equation.
x=-5
The equation is now solved. Solutions are the same.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}