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4a^{2}-4a-3=0
Subtract 3 from both sides.
a+b=-4 ab=4\left(-3\right)=-12
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4a^{2}+aa+ba-3. To find a and b, set up a system to be solved.
1,-12 2,-6 3,-4
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 -12.
1-12=-11 2-6=-4 3-4=-1
Calculate the sum for each pair.
a=-6 b=2
The solution is the pair that gives sum -4.
\left(4a^{2}-6a\right)+\left(2a-3\right)
Rewrite 4a^{2}-4a-3 as \left(4a^{2}-6a\right)+\left(2a-3\right).
2a\left(2a-3\right)+2a-3
Factor out 2a in 4a^{2}-6a.
\left(2a-3\right)\left(2a+1\right)
Factor out common term 2a-3 by using distributive property.
a=\frac{3}{2} a=-\frac{1}{2}
To find equation solutions, solve 2a-3=0 and 2a+1=0.
4a^{2}-4a=3
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.
4a^{2}-4a-3=3-3
Subtract 3 from both sides of the equation.
4a^{2}-4a-3=0
Subtracting 3 from itself leaves 0.
a=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\left(-3\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -4 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-3\right)}}{2\times 4}
Square -4.
a=\frac{-\left(-4\right)±\sqrt{16-16\left(-3\right)}}{2\times 4}
Multiply -4 times 4.
a=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 4}
Multiply -16 times -3.
a=\frac{-\left(-4\right)±\sqrt{64}}{2\times 4}
Add 16 to 48.
a=\frac{-\left(-4\right)±8}{2\times 4}
Take the square root of 64.
a=\frac{4±8}{2\times 4}
The opposite of -4 is 4.
a=\frac{4±8}{8}
Multiply 2 times 4.
a=\frac{12}{8}
Now solve the equation a=\frac{4±8}{8} when ± is plus. Add 4 to 8.
a=\frac{3}{2}
Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.
a=-\frac{4}{8}
Now solve the equation a=\frac{4±8}{8} when ± is minus. Subtract 8 from 4.
a=-\frac{1}{2}
Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.
a=\frac{3}{2} a=-\frac{1}{2}
The equation is now solved.
4a^{2}-4a=3
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{4a^{2}-4a}{4}=\frac{3}{4}
Divide both sides by 4.
a^{2}+\left(-\frac{4}{4}\right)a=\frac{3}{4}
Dividing by 4 undoes the multiplication by 4.
a^{2}-a=\frac{3}{4}
Divide -4 by 4.
a^{2}-a+\left(-\frac{1}{2}\right)^{2}=\frac{3}{4}+\left(-\frac{1}{2}\right)^{2}
Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-a+\frac{1}{4}=\frac{3+1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
a^{2}-a+\frac{1}{4}=1
Add \frac{3}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a-\frac{1}{2}\right)^{2}=1
Factor a^{2}-a+\frac{1}{4}. 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(a-\frac{1}{2}\right)^{2}}=\sqrt{1}
Take the square root of both sides of the equation.
a-\frac{1}{2}=1 a-\frac{1}{2}=-1
Simplify.
a=\frac{3}{2} a=-\frac{1}{2}
Add \frac{1}{2} to both sides of the equation.