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2x^{2}-5x+3=0
Divide both sides by 2.
a+b=-5 ab=2\times 3=6
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+3. To find a and b, set up a system to be solved.
-1,-6 -2,-3
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 6.
-1-6=-7 -2-3=-5
Calculate the sum for each pair.
a=-3 b=-2
The solution is the pair that gives sum -5.
\left(2x^{2}-3x\right)+\left(-2x+3\right)
Rewrite 2x^{2}-5x+3 as \left(2x^{2}-3x\right)+\left(-2x+3\right).
x\left(2x-3\right)-\left(2x-3\right)
Factor out x in the first and -1 in the second group.
\left(2x-3\right)\left(x-1\right)
Factor out common term 2x-3 by using distributive property.
x=\frac{3}{2} x=1
To find equation solutions, solve 2x-3=0 and x-1=0.
4x^{2}-10x+6=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 4\times 6}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -10 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-10\right)±\sqrt{100-4\times 4\times 6}}{2\times 4}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-16\times 6}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times 4}
Multiply -16 times 6.
x=\frac{-\left(-10\right)±\sqrt{4}}{2\times 4}
Add 100 to -96.
x=\frac{-\left(-10\right)±2}{2\times 4}
Take the square root of 4.
x=\frac{10±2}{2\times 4}
The opposite of -10 is 10.
x=\frac{10±2}{8}
Multiply 2 times 4.
x=\frac{12}{8}
Now solve the equation x=\frac{10±2}{8} when ± is plus. Add 10 to 2.
x=\frac{3}{2}
Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.
x=\frac{8}{8}
Now solve the equation x=\frac{10±2}{8} when ± is minus. Subtract 2 from 10.
x=1
Divide 8 by 8.
x=\frac{3}{2} x=1
The equation is now solved.
4x^{2}-10x+6=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.
4x^{2}-10x+6-6=-6
Subtract 6 from both sides of the equation.
4x^{2}-10x=-6
Subtracting 6 from itself leaves 0.
\frac{4x^{2}-10x}{4}=-\frac{6}{4}
Divide both sides by 4.
x^{2}+\left(-\frac{10}{4}\right)x=-\frac{6}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}-\frac{5}{2}x=-\frac{6}{4}
Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{5}{2}x=-\frac{3}{2}
Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-\frac{3}{2}+\left(-\frac{5}{4}\right)^{2}
Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{3}{2}+\frac{25}{16}
Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{1}{16}
Add -\frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{4}\right)^{2}=\frac{1}{16}
Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. 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-\frac{5}{4}\right)^{2}}=\sqrt{\frac{1}{16}}
Take the square root of both sides of the equation.
x-\frac{5}{4}=\frac{1}{4} x-\frac{5}{4}=-\frac{1}{4}
Simplify.
x=\frac{3}{2} x=1
Add \frac{5}{4} to both sides of the equation.
x ^ 2 -\frac{5}{2}x +\frac{3}{2} = 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 4
r + s = \frac{5}{2} rs = \frac{3}{2}
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 = \frac{5}{4} - u s = \frac{5}{4} + u
Two numbers r and s sum up to \frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{2} = \frac{5}{4}. 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>
(\frac{5}{4} - u) (\frac{5}{4} + u) = \frac{3}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{2}
\frac{25}{16} - u^2 = \frac{3}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{2}-\frac{25}{16} = -\frac{1}{16}
Simplify the expression by subtracting \frac{25}{16} on both sides
u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{4} - \frac{1}{4} = 1 s = \frac{5}{4} + \frac{1}{4} = 1.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.