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a+b=-24 ab=28\times 5=140
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 28x^{2}+ax+bx+5. To find a and b, set up a system to be solved.
-1,-140 -2,-70 -4,-35 -5,-28 -7,-20 -10,-14
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 140.
-1-140=-141 -2-70=-72 -4-35=-39 -5-28=-33 -7-20=-27 -10-14=-24
Calculate the sum for each pair.
a=-14 b=-10
The solution is the pair that gives sum -24.
\left(28x^{2}-14x\right)+\left(-10x+5\right)
Rewrite 28x^{2}-24x+5 as \left(28x^{2}-14x\right)+\left(-10x+5\right).
14x\left(2x-1\right)-5\left(2x-1\right)
Factor out 14x in the first and -5 in the second group.
\left(2x-1\right)\left(14x-5\right)
Factor out common term 2x-1 by using distributive property.
x=\frac{1}{2} x=\frac{5}{14}
To find equation solutions, solve 2x-1=0 and 14x-5=0.
28x^{2}-24x+5=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(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 28\times 5}}{2\times 28}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 28 for a, -24 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-24\right)±\sqrt{576-4\times 28\times 5}}{2\times 28}
Square -24.
x=\frac{-\left(-24\right)±\sqrt{576-112\times 5}}{2\times 28}
Multiply -4 times 28.
x=\frac{-\left(-24\right)±\sqrt{576-560}}{2\times 28}
Multiply -112 times 5.
x=\frac{-\left(-24\right)±\sqrt{16}}{2\times 28}
Add 576 to -560.
x=\frac{-\left(-24\right)±4}{2\times 28}
Take the square root of 16.
x=\frac{24±4}{2\times 28}
The opposite of -24 is 24.
x=\frac{24±4}{56}
Multiply 2 times 28.
x=\frac{28}{56}
Now solve the equation x=\frac{24±4}{56} when ± is plus. Add 24 to 4.
x=\frac{1}{2}
Reduce the fraction \frac{28}{56} to lowest terms by extracting and canceling out 28.
x=\frac{20}{56}
Now solve the equation x=\frac{24±4}{56} when ± is minus. Subtract 4 from 24.
x=\frac{5}{14}
Reduce the fraction \frac{20}{56} to lowest terms by extracting and canceling out 4.
x=\frac{1}{2} x=\frac{5}{14}
The equation is now solved.
28x^{2}-24x+5=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.
28x^{2}-24x+5-5=-5
Subtract 5 from both sides of the equation.
28x^{2}-24x=-5
Subtracting 5 from itself leaves 0.
\frac{28x^{2}-24x}{28}=-\frac{5}{28}
Divide both sides by 28.
x^{2}+\left(-\frac{24}{28}\right)x=-\frac{5}{28}
Dividing by 28 undoes the multiplication by 28.
x^{2}-\frac{6}{7}x=-\frac{5}{28}
Reduce the fraction \frac{-24}{28} to lowest terms by extracting and canceling out 4.
x^{2}-\frac{6}{7}x+\left(-\frac{3}{7}\right)^{2}=-\frac{5}{28}+\left(-\frac{3}{7}\right)^{2}
Divide -\frac{6}{7}, the coefficient of the x term, by 2 to get -\frac{3}{7}. Then add the square of -\frac{3}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{6}{7}x+\frac{9}{49}=-\frac{5}{28}+\frac{9}{49}
Square -\frac{3}{7} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{6}{7}x+\frac{9}{49}=\frac{1}{196}
Add -\frac{5}{28} to \frac{9}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{7}\right)^{2}=\frac{1}{196}
Factor x^{2}-\frac{6}{7}x+\frac{9}{49}. 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{3}{7}\right)^{2}}=\sqrt{\frac{1}{196}}
Take the square root of both sides of the equation.
x-\frac{3}{7}=\frac{1}{14} x-\frac{3}{7}=-\frac{1}{14}
Simplify.
x=\frac{1}{2} x=\frac{5}{14}
Add \frac{3}{7} to both sides of the equation.
x ^ 2 -\frac{6}{7}x +\frac{5}{28} = 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 28
r + s = \frac{6}{7} rs = \frac{5}{28}
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{3}{7} - u s = \frac{3}{7} + u
Two numbers r and s sum up to \frac{6}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{6}{7} = \frac{3}{7}. 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{3}{7} - u) (\frac{3}{7} + u) = \frac{5}{28}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{28}
\frac{9}{49} - u^2 = \frac{5}{28}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{5}{28}-\frac{9}{49} = -\frac{1}{196}
Simplify the expression by subtracting \frac{9}{49} on both sides
u^2 = \frac{1}{196} u = \pm\sqrt{\frac{1}{196}} = \pm \frac{1}{14}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{3}{7} - \frac{1}{14} = 0.357 s = \frac{3}{7} + \frac{1}{14} = 0.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.