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32x-16x^{2}-12=0
Subtract 12 from both sides.
8x-4x^{2}-3=0
Divide both sides by 4.
-4x^{2}+8x-3=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=8 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 -4x^{2}+ax+bx-3. To find a and b, set up a system to be solved.
1,12 2,6 3,4
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.
1+12=13 2+6=8 3+4=7
Calculate the sum for each pair.
a=6 b=2
The solution is the pair that gives sum 8.
\left(-4x^{2}+6x\right)+\left(2x-3\right)
Rewrite -4x^{2}+8x-3 as \left(-4x^{2}+6x\right)+\left(2x-3\right).
-2x\left(2x-3\right)+2x-3
Factor out -2x in -4x^{2}+6x.
\left(2x-3\right)\left(-2x+1\right)
Factor out common term 2x-3 by using distributive property.
x=\frac{3}{2} x=\frac{1}{2}
To find equation solutions, solve 2x-3=0 and -2x+1=0.
-16x^{2}+32x=12
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.
-16x^{2}+32x-12=12-12
Subtract 12 from both sides of the equation.
-16x^{2}+32x-12=0
Subtracting 12 from itself leaves 0.
x=\frac{-32±\sqrt{32^{2}-4\left(-16\right)\left(-12\right)}}{2\left(-16\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 32 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-32±\sqrt{1024-4\left(-16\right)\left(-12\right)}}{2\left(-16\right)}
Square 32.
x=\frac{-32±\sqrt{1024+64\left(-12\right)}}{2\left(-16\right)}
Multiply -4 times -16.
x=\frac{-32±\sqrt{1024-768}}{2\left(-16\right)}
Multiply 64 times -12.
x=\frac{-32±\sqrt{256}}{2\left(-16\right)}
Add 1024 to -768.
x=\frac{-32±16}{2\left(-16\right)}
Take the square root of 256.
x=\frac{-32±16}{-32}
Multiply 2 times -16.
x=-\frac{16}{-32}
Now solve the equation x=\frac{-32±16}{-32} when ± is plus. Add -32 to 16.
x=\frac{1}{2}
Reduce the fraction \frac{-16}{-32} to lowest terms by extracting and canceling out 16.
x=-\frac{48}{-32}
Now solve the equation x=\frac{-32±16}{-32} when ± is minus. Subtract 16 from -32.
x=\frac{3}{2}
Reduce the fraction \frac{-48}{-32} to lowest terms by extracting and canceling out 16.
x=\frac{1}{2} x=\frac{3}{2}
The equation is now solved.
-16x^{2}+32x=12
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{-16x^{2}+32x}{-16}=\frac{12}{-16}
Divide both sides by -16.
x^{2}+\frac{32}{-16}x=\frac{12}{-16}
Dividing by -16 undoes the multiplication by -16.
x^{2}-2x=\frac{12}{-16}
Divide 32 by -16.
x^{2}-2x=-\frac{3}{4}
Reduce the fraction \frac{12}{-16} to lowest terms by extracting and canceling out 4.
x^{2}-2x+1=-\frac{3}{4}+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-2x+1=\frac{1}{4}
Add -\frac{3}{4} to 1.
\left(x-1\right)^{2}=\frac{1}{4}
Factor x^{2}-2x+1. 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-1\right)^{2}}=\sqrt{\frac{1}{4}}
Take the square root of both sides of the equation.
x-1=\frac{1}{2} x-1=-\frac{1}{2}
Simplify.
x=\frac{3}{2} x=\frac{1}{2}
Add 1 to both sides of the equation.