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4u^{2}-5u-6=0
Subtract 6 from both sides.
a+b=-5 ab=4\left(-6\right)=-24
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4u^{2}+au+bu-6. To find a and b, set up a system to be solved.
1,-24 2,-12 3,-8 4,-6
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 -24.
1-24=-23 2-12=-10 3-8=-5 4-6=-2
Calculate the sum for each pair.
a=-8 b=3
The solution is the pair that gives sum -5.
\left(4u^{2}-8u\right)+\left(3u-6\right)
Rewrite 4u^{2}-5u-6 as \left(4u^{2}-8u\right)+\left(3u-6\right).
4u\left(u-2\right)+3\left(u-2\right)
Factor out 4u in the first and 3 in the second group.
\left(u-2\right)\left(4u+3\right)
Factor out common term u-2 by using distributive property.
u=2 u=-\frac{3}{4}
To find equation solutions, solve u-2=0 and 4u+3=0.
4u^{2}-5u=6
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.
4u^{2}-5u-6=6-6
Subtract 6 from both sides of the equation.
4u^{2}-5u-6=0
Subtracting 6 from itself leaves 0.
u=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\left(-6\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
u=\frac{-\left(-5\right)±\sqrt{25-4\times 4\left(-6\right)}}{2\times 4}
Square -5.
u=\frac{-\left(-5\right)±\sqrt{25-16\left(-6\right)}}{2\times 4}
Multiply -4 times 4.
u=\frac{-\left(-5\right)±\sqrt{25+96}}{2\times 4}
Multiply -16 times -6.
u=\frac{-\left(-5\right)±\sqrt{121}}{2\times 4}
Add 25 to 96.
u=\frac{-\left(-5\right)±11}{2\times 4}
Take the square root of 121.
u=\frac{5±11}{2\times 4}
The opposite of -5 is 5.
u=\frac{5±11}{8}
Multiply 2 times 4.
u=\frac{16}{8}
Now solve the equation u=\frac{5±11}{8} when ± is plus. Add 5 to 11.
u=2
Divide 16 by 8.
u=-\frac{6}{8}
Now solve the equation u=\frac{5±11}{8} when ± is minus. Subtract 11 from 5.
u=-\frac{3}{4}
Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.
u=2 u=-\frac{3}{4}
The equation is now solved.
4u^{2}-5u=6
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{4u^{2}-5u}{4}=\frac{6}{4}
Divide both sides by 4.
u^{2}-\frac{5}{4}u=\frac{6}{4}
Dividing by 4 undoes the multiplication by 4.
u^{2}-\frac{5}{4}u=\frac{3}{2}
Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.
u^{2}-\frac{5}{4}u+\left(-\frac{5}{8}\right)^{2}=\frac{3}{2}+\left(-\frac{5}{8}\right)^{2}
Divide -\frac{5}{4}, the coefficient of the x term, by 2 to get -\frac{5}{8}. Then add the square of -\frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
u^{2}-\frac{5}{4}u+\frac{25}{64}=\frac{3}{2}+\frac{25}{64}
Square -\frac{5}{8} by squaring both the numerator and the denominator of the fraction.
u^{2}-\frac{5}{4}u+\frac{25}{64}=\frac{121}{64}
Add \frac{3}{2} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(u-\frac{5}{8}\right)^{2}=\frac{121}{64}
Factor u^{2}-\frac{5}{4}u+\frac{25}{64}. 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{5}{8}\right)^{2}}=\sqrt{\frac{121}{64}}
Take the square root of both sides of the equation.
u-\frac{5}{8}=\frac{11}{8} u-\frac{5}{8}=-\frac{11}{8}
Simplify.
u=2 u=-\frac{3}{4}
Add \frac{5}{8} to both sides of the equation.