Skip to main content
Solve for x
Tick mark Image
Graph

Similar Problems from Web Search

Share

a+b=-17 ab=6\times 10=60
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx+10. To find a and b, set up a system to be solved.
-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10
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 60.
-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16
Calculate the sum for each pair.
a=-12 b=-5
The solution is the pair that gives sum -17.
\left(6x^{2}-12x\right)+\left(-5x+10\right)
Rewrite 6x^{2}-17x+10 as \left(6x^{2}-12x\right)+\left(-5x+10\right).
6x\left(x-2\right)-5\left(x-2\right)
Factor out 6x in the first and -5 in the second group.
\left(x-2\right)\left(6x-5\right)
Factor out common term x-2 by using distributive property.
x=2 x=\frac{5}{6}
To find equation solutions, solve x-2=0 and 6x-5=0.
6x^{2}-17x+10=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(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 6\times 10}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -17 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-17\right)±\sqrt{289-4\times 6\times 10}}{2\times 6}
Square -17.
x=\frac{-\left(-17\right)±\sqrt{289-24\times 10}}{2\times 6}
Multiply -4 times 6.
x=\frac{-\left(-17\right)±\sqrt{289-240}}{2\times 6}
Multiply -24 times 10.
x=\frac{-\left(-17\right)±\sqrt{49}}{2\times 6}
Add 289 to -240.
x=\frac{-\left(-17\right)±7}{2\times 6}
Take the square root of 49.
x=\frac{17±7}{2\times 6}
The opposite of -17 is 17.
x=\frac{17±7}{12}
Multiply 2 times 6.
x=\frac{24}{12}
Now solve the equation x=\frac{17±7}{12} when ± is plus. Add 17 to 7.
x=2
Divide 24 by 12.
x=\frac{10}{12}
Now solve the equation x=\frac{17±7}{12} when ± is minus. Subtract 7 from 17.
x=\frac{5}{6}
Reduce the fraction \frac{10}{12} to lowest terms by extracting and canceling out 2.
x=2 x=\frac{5}{6}
The equation is now solved.
6x^{2}-17x+10=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.
6x^{2}-17x+10-10=-10
Subtract 10 from both sides of the equation.
6x^{2}-17x=-10
Subtracting 10 from itself leaves 0.
\frac{6x^{2}-17x}{6}=-\frac{10}{6}
Divide both sides by 6.
x^{2}-\frac{17}{6}x=-\frac{10}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}-\frac{17}{6}x=-\frac{5}{3}
Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{17}{6}x+\left(-\frac{17}{12}\right)^{2}=-\frac{5}{3}+\left(-\frac{17}{12}\right)^{2}
Divide -\frac{17}{6}, the coefficient of the x term, by 2 to get -\frac{17}{12}. Then add the square of -\frac{17}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{17}{6}x+\frac{289}{144}=-\frac{5}{3}+\frac{289}{144}
Square -\frac{17}{12} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{17}{6}x+\frac{289}{144}=\frac{49}{144}
Add -\frac{5}{3} to \frac{289}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{17}{12}\right)^{2}=\frac{49}{144}
Factor x^{2}-\frac{17}{6}x+\frac{289}{144}. 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{17}{12}\right)^{2}}=\sqrt{\frac{49}{144}}
Take the square root of both sides of the equation.
x-\frac{17}{12}=\frac{7}{12} x-\frac{17}{12}=-\frac{7}{12}
Simplify.
x=2 x=\frac{5}{6}
Add \frac{17}{12} to both sides of the equation.