Solve for x
x=\sqrt{6}+1\approx 3.449489743
x=1-\sqrt{6}\approx -1.449489743
Graph
Share
Copied to clipboard
5x^{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 5\left(-25\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 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 5\left(-25\right)}}{2\times 5}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-20\left(-25\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-10\right)±\sqrt{100+500}}{2\times 5}
Multiply -20 times -25.
x=\frac{-\left(-10\right)±\sqrt{600}}{2\times 5}
Add 100 to 500.
x=\frac{-\left(-10\right)±10\sqrt{6}}{2\times 5}
Take the square root of 600.
x=\frac{10±10\sqrt{6}}{2\times 5}
The opposite of -10 is 10.
x=\frac{10±10\sqrt{6}}{10}
Multiply 2 times 5.
x=\frac{10\sqrt{6}+10}{10}
Now solve the equation x=\frac{10±10\sqrt{6}}{10} when ± is plus. Add 10 to 10\sqrt{6}.
x=\sqrt{6}+1
Divide 10+10\sqrt{6} by 10.
x=\frac{10-10\sqrt{6}}{10}
Now solve the equation x=\frac{10±10\sqrt{6}}{10} when ± is minus. Subtract 10\sqrt{6} from 10.
x=1-\sqrt{6}
Divide 10-10\sqrt{6} by 10.
x=\sqrt{6}+1 x=1-\sqrt{6}
The equation is now solved.
5x^{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.
5x^{2}-10x-25-\left(-25\right)=-\left(-25\right)
Add 25 to both sides of the equation.
5x^{2}-10x=-\left(-25\right)
Subtracting -25 from itself leaves 0.
5x^{2}-10x=25
Subtract -25 from 0.
\frac{5x^{2}-10x}{5}=\frac{25}{5}
Divide both sides by 5.
x^{2}+\left(-\frac{10}{5}\right)x=\frac{25}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-2x=\frac{25}{5}
Divide -10 by 5.
x^{2}-2x=5
Divide 25 by 5.
x^{2}-2x+1=5+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.
x^{2}-2x+1=6
Add 5 to 1.
\left(x-1\right)^{2}=6
Factor x^{2}-2x+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(x-1\right)^{2}}=\sqrt{6}
Take the square root of both sides of the equation.
x-1=\sqrt{6} x-1=-\sqrt{6}
Simplify.
x=\sqrt{6}+1 x=1-\sqrt{6}
Add 1 to both sides of the equation.
x ^ 2 -2x -5 = 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 = -5
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) = -5
To solve for unknown quantity u, substitute these in the product equation rs = -5
1 - u^2 = -5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -5-1 = -6
Simplify the expression by subtracting 1 on both sides
u^2 = 6 u = \pm\sqrt{6} = \pm \sqrt{6}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - \sqrt{6} = -1.449 s = 1 + \sqrt{6} = 3.449
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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}