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3u^{2}-14u-5=0
Subtract 5 from both sides.
a+b=-14 ab=3\left(-5\right)=-15
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3u^{2}+au+bu-5. To find a and b, set up a system to be solved.
1,-15 3,-5
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -15.
1-15=-14 3-5=-2
Calculate the sum for each pair.
a=-15 b=1
The solution is the pair that gives sum -14.
\left(3u^{2}-15u\right)+\left(u-5\right)
Rewrite 3u^{2}-14u-5 as \left(3u^{2}-15u\right)+\left(u-5\right).
3u\left(u-5\right)+u-5
Factor out 3u in 3u^{2}-15u.
\left(u-5\right)\left(3u+1\right)
Factor out common term u-5 by using distributive property.
u=5 u=-\frac{1}{3}
To find equation solutions, solve u-5=0 and 3u+1=0.
3u^{2}-14u=5
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.
3u^{2}-14u-5=5-5
Subtract 5 from both sides of the equation.
3u^{2}-14u-5=0
Subtracting 5 from itself leaves 0.
u=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 3\left(-5\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -14 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
u=\frac{-\left(-14\right)±\sqrt{196-4\times 3\left(-5\right)}}{2\times 3}
Square -14.
u=\frac{-\left(-14\right)±\sqrt{196-12\left(-5\right)}}{2\times 3}
Multiply -4 times 3.
u=\frac{-\left(-14\right)±\sqrt{196+60}}{2\times 3}
Multiply -12 times -5.
u=\frac{-\left(-14\right)±\sqrt{256}}{2\times 3}
Add 196 to 60.
u=\frac{-\left(-14\right)±16}{2\times 3}
Take the square root of 256.
u=\frac{14±16}{2\times 3}
The opposite of -14 is 14.
u=\frac{14±16}{6}
Multiply 2 times 3.
u=\frac{30}{6}
Now solve the equation u=\frac{14±16}{6} when ± is plus. Add 14 to 16.
u=5
Divide 30 by 6.
u=-\frac{2}{6}
Now solve the equation u=\frac{14±16}{6} when ± is minus. Subtract 16 from 14.
u=-\frac{1}{3}
Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.
u=5 u=-\frac{1}{3}
The equation is now solved.
3u^{2}-14u=5
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.
\frac{3u^{2}-14u}{3}=\frac{5}{3}
Divide both sides by 3.
u^{2}-\frac{14}{3}u=\frac{5}{3}
Dividing by 3 undoes the multiplication by 3.
u^{2}-\frac{14}{3}u+\left(-\frac{7}{3}\right)^{2}=\frac{5}{3}+\left(-\frac{7}{3}\right)^{2}
Divide -\frac{14}{3}, the coefficient of the x term, by 2 to get -\frac{7}{3}. Then add the square of -\frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
u^{2}-\frac{14}{3}u+\frac{49}{9}=\frac{5}{3}+\frac{49}{9}
Square -\frac{7}{3} by squaring both the numerator and the denominator of the fraction.
u^{2}-\frac{14}{3}u+\frac{49}{9}=\frac{64}{9}
Add \frac{5}{3} to \frac{49}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(u-\frac{7}{3}\right)^{2}=\frac{64}{9}
Factor u^{2}-\frac{14}{3}u+\frac{49}{9}. 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(u-\frac{7}{3}\right)^{2}}=\sqrt{\frac{64}{9}}
Take the square root of both sides of the equation.
u-\frac{7}{3}=\frac{8}{3} u-\frac{7}{3}=-\frac{8}{3}
Simplify.
u=5 u=-\frac{1}{3}
Add \frac{7}{3} to both sides of the equation.