Solve for x (complex solution)
x=\sqrt{2}-5\approx -3.585786438
x=-\left(\sqrt{2}+5\right)\approx -6.414213562
Solve for x
x=\sqrt{2}-5\approx -3.585786438
x=-\sqrt{2}-5\approx -6.414213562
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x^{2}+10x+23=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{-10±\sqrt{10^{2}-4\times 23}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 23 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-10±\sqrt{100-4\times 23}}{2}
Square 10.
x=\frac{-10±\sqrt{100-92}}{2}
Multiply -4 times 23.
x=\frac{-10±\sqrt{8}}{2}
Add 100 to -92.
x=\frac{-10±2\sqrt{2}}{2}
Take the square root of 8.
x=\frac{2\sqrt{2}-10}{2}
Now solve the equation x=\frac{-10±2\sqrt{2}}{2} when ± is plus. Add -10 to 2\sqrt{2}.
x=\sqrt{2}-5
Divide -10+2\sqrt{2} by 2.
x=\frac{-2\sqrt{2}-10}{2}
Now solve the equation x=\frac{-10±2\sqrt{2}}{2} when ± is minus. Subtract 2\sqrt{2} from -10.
x=-\sqrt{2}-5
Divide -10-2\sqrt{2} by 2.
x=\sqrt{2}-5 x=-\sqrt{2}-5
The equation is now solved.
x^{2}+10x+23=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}+10x+23-23=-23
Subtract 23 from both sides of the equation.
x^{2}+10x=-23
Subtracting 23 from itself leaves 0.
x^{2}+10x+5^{2}=-23+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=-23+25
Square 5.
x^{2}+10x+25=2
Add -23 to 25.
\left(x+5\right)^{2}=2
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{2}
Take the square root of both sides of the equation.
x+5=\sqrt{2} x+5=-\sqrt{2}
Simplify.
x=\sqrt{2}-5 x=-\sqrt{2}-5
Subtract 5 from both sides of the equation.
x ^ 2 +10x +23 = 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 = 23
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) = 23
To solve for unknown quantity u, substitute these in the product equation rs = 23
25 - u^2 = 23
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 23-25 = -2
Simplify the expression by subtracting 25 on both sides
u^2 = 2 u = \pm\sqrt{2} = \pm \sqrt{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-5 - \sqrt{2} = -6.414 s = -5 + \sqrt{2} = -3.586
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
x^{2}+10x+23=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{-10±\sqrt{10^{2}-4\times 23}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 23 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-10±\sqrt{100-4\times 23}}{2}
Square 10.
x=\frac{-10±\sqrt{100-92}}{2}
Multiply -4 times 23.
x=\frac{-10±\sqrt{8}}{2}
Add 100 to -92.
x=\frac{-10±2\sqrt{2}}{2}
Take the square root of 8.
x=\frac{2\sqrt{2}-10}{2}
Now solve the equation x=\frac{-10±2\sqrt{2}}{2} when ± is plus. Add -10 to 2\sqrt{2}.
x=\sqrt{2}-5
Divide -10+2\sqrt{2} by 2.
x=\frac{-2\sqrt{2}-10}{2}
Now solve the equation x=\frac{-10±2\sqrt{2}}{2} when ± is minus. Subtract 2\sqrt{2} from -10.
x=-\sqrt{2}-5
Divide -10-2\sqrt{2} by 2.
x=\sqrt{2}-5 x=-\sqrt{2}-5
The equation is now solved.
x^{2}+10x+23=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}+10x+23-23=-23
Subtract 23 from both sides of the equation.
x^{2}+10x=-23
Subtracting 23 from itself leaves 0.
x^{2}+10x+5^{2}=-23+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=-23+25
Square 5.
x^{2}+10x+25=2
Add -23 to 25.
\left(x+5\right)^{2}=2
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{2}
Take the square root of both sides of the equation.
x+5=\sqrt{2} x+5=-\sqrt{2}
Simplify.
x=\sqrt{2}-5 x=-\sqrt{2}-5
Subtract 5 from both sides of the equation.
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}