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a+b=-38 ab=69\times 5=345
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 69x^{2}+ax+bx+5. To find a and b, set up a system to be solved.
-1,-345 -3,-115 -5,-69 -15,-23
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 345.
-1-345=-346 -3-115=-118 -5-69=-74 -15-23=-38
Calculate the sum for each pair.
a=-23 b=-15
The solution is the pair that gives sum -38.
\left(69x^{2}-23x\right)+\left(-15x+5\right)
Rewrite 69x^{2}-38x+5 as \left(69x^{2}-23x\right)+\left(-15x+5\right).
23x\left(3x-1\right)-5\left(3x-1\right)
Factor out 23x in the first and -5 in the second group.
\left(3x-1\right)\left(23x-5\right)
Factor out common term 3x-1 by using distributive property.
x=\frac{1}{3} x=\frac{5}{23}
To find equation solutions, solve 3x-1=0 and 23x-5=0.
69x^{2}-38x+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(-38\right)±\sqrt{\left(-38\right)^{2}-4\times 69\times 5}}{2\times 69}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 69 for a, -38 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-38\right)±\sqrt{1444-4\times 69\times 5}}{2\times 69}
Square -38.
x=\frac{-\left(-38\right)±\sqrt{1444-276\times 5}}{2\times 69}
Multiply -4 times 69.
x=\frac{-\left(-38\right)±\sqrt{1444-1380}}{2\times 69}
Multiply -276 times 5.
x=\frac{-\left(-38\right)±\sqrt{64}}{2\times 69}
Add 1444 to -1380.
x=\frac{-\left(-38\right)±8}{2\times 69}
Take the square root of 64.
x=\frac{38±8}{2\times 69}
The opposite of -38 is 38.
x=\frac{38±8}{138}
Multiply 2 times 69.
x=\frac{46}{138}
Now solve the equation x=\frac{38±8}{138} when ± is plus. Add 38 to 8.
x=\frac{1}{3}
Reduce the fraction \frac{46}{138} to lowest terms by extracting and canceling out 46.
x=\frac{30}{138}
Now solve the equation x=\frac{38±8}{138} when ± is minus. Subtract 8 from 38.
x=\frac{5}{23}
Reduce the fraction \frac{30}{138} to lowest terms by extracting and canceling out 6.
x=\frac{1}{3} x=\frac{5}{23}
The equation is now solved.
69x^{2}-38x+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.
69x^{2}-38x+5-5=-5
Subtract 5 from both sides of the equation.
69x^{2}-38x=-5
Subtracting 5 from itself leaves 0.
\frac{69x^{2}-38x}{69}=-\frac{5}{69}
Divide both sides by 69.
x^{2}-\frac{38}{69}x=-\frac{5}{69}
Dividing by 69 undoes the multiplication by 69.
x^{2}-\frac{38}{69}x+\left(-\frac{19}{69}\right)^{2}=-\frac{5}{69}+\left(-\frac{19}{69}\right)^{2}
Divide -\frac{38}{69}, the coefficient of the x term, by 2 to get -\frac{19}{69}. Then add the square of -\frac{19}{69} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{38}{69}x+\frac{361}{4761}=-\frac{5}{69}+\frac{361}{4761}
Square -\frac{19}{69} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{38}{69}x+\frac{361}{4761}=\frac{16}{4761}
Add -\frac{5}{69} to \frac{361}{4761} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{19}{69}\right)^{2}=\frac{16}{4761}
Factor x^{2}-\frac{38}{69}x+\frac{361}{4761}. 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{19}{69}\right)^{2}}=\sqrt{\frac{16}{4761}}
Take the square root of both sides of the equation.
x-\frac{19}{69}=\frac{4}{69} x-\frac{19}{69}=-\frac{4}{69}
Simplify.
x=\frac{1}{3} x=\frac{5}{23}
Add \frac{19}{69} to both sides of the equation.