Skip to main content
Solve for x
Tick mark Image
Graph

Similar Problems from Web Search

Share

5x^{2}-10x+3=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\times 3}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -10 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-10\right)±\sqrt{100-4\times 5\times 3}}{2\times 5}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-20\times 3}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-10\right)±\sqrt{100-60}}{2\times 5}
Multiply -20 times 3.
x=\frac{-\left(-10\right)±\sqrt{40}}{2\times 5}
Add 100 to -60.
x=\frac{-\left(-10\right)±2\sqrt{10}}{2\times 5}
Take the square root of 40.
x=\frac{10±2\sqrt{10}}{2\times 5}
The opposite of -10 is 10.
x=\frac{10±2\sqrt{10}}{10}
Multiply 2 times 5.
x=\frac{2\sqrt{10}+10}{10}
Now solve the equation x=\frac{10±2\sqrt{10}}{10} when ± is plus. Add 10 to 2\sqrt{10}.
x=\frac{\sqrt{10}}{5}+1
Divide 10+2\sqrt{10} by 10.
x=\frac{10-2\sqrt{10}}{10}
Now solve the equation x=\frac{10±2\sqrt{10}}{10} when ± is minus. Subtract 2\sqrt{10} from 10.
x=-\frac{\sqrt{10}}{5}+1
Divide 10-2\sqrt{10} by 10.
x=\frac{\sqrt{10}}{5}+1 x=-\frac{\sqrt{10}}{5}+1
The equation is now solved.
5x^{2}-10x+3=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+3-3=-3
Subtract 3 from both sides of the equation.
5x^{2}-10x=-3
Subtracting 3 from itself leaves 0.
\frac{5x^{2}-10x}{5}=-\frac{3}{5}
Divide both sides by 5.
x^{2}+\left(-\frac{10}{5}\right)x=-\frac{3}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-2x=-\frac{3}{5}
Divide -10 by 5.
x^{2}-2x+1=-\frac{3}{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=\frac{2}{5}
Add -\frac{3}{5} to 1.
\left(x-1\right)^{2}=\frac{2}{5}
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{\frac{2}{5}}
Take the square root of both sides of the equation.
x-1=\frac{\sqrt{10}}{5} x-1=-\frac{\sqrt{10}}{5}
Simplify.
x=\frac{\sqrt{10}}{5}+1 x=-\frac{\sqrt{10}}{5}+1
Add 1 to both sides of the equation.
x ^ 2 -2x +\frac{3}{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 = \frac{3}{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) = \frac{3}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{5}
1 - u^2 = \frac{3}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{5}-1 = -\frac{2}{5}
Simplify the expression by subtracting 1 on both sides
u^2 = \frac{2}{5} u = \pm\sqrt{\frac{2}{5}} = \pm \frac{\sqrt{2}}{\sqrt{5}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - \frac{\sqrt{2}}{\sqrt{5}} = 0.368 s = 1 + \frac{\sqrt{2}}{\sqrt{5}} = 1.632
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.